I often teach a short unit on induction in elementary calculus classes—it comes up very naturally when trying to prove a version of the power rule:

Lemma 1: (Product Rule) Let $f$ and $g$ be real valued functions on $\mathbb{R}$. At any point $x$ where both $f$ and $g$ are differentiable, $$ (fg)'(x) = f'(x) g(x) + f(x) g'(x). $$

As it is not really germane to this discussion, I'll assume that the product rule has already been proved.

Theorem 2: (Power Rule, version 1) For any natural number $n$, $$\frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}.$$

Proof of Theorem 2 (by induction): As a basis for induction, note that $$ \frac{\mathrm{d}}{\mathrm{d}x} x^1 = \lim_{h\to 0} \frac{(x+h)^1 - x^1}{h} = \lim_{h\to 0} \frac{h}{h} = 1 = x^0. $$ Suppose that there is some natural number $k$ such that $$ \frac{\mathrm{d}}{\mathrm{d}x} x^k = k x^{k-1}. $$ Then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} x^{k+1} &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ x \cdot x^k \right] \\ &= \left[ \frac{\mathrm{d}}{\mathrm{d}x} x \right] x^k + x \left[ \frac{\mathrm{d}}{\mathrm{d}x} x^k \right] && \text{(Product Rule)} \\ &= 1 \cdot x^k + k x \cdot x^{k-1} && \text{(Induction Hypothesis)} \\ &= (k+1) x^k, \end{align} which is the desired result. $\square$

It is worth noting, perhaps, that most elementary calculus texts prove this version of the power rule by appealing to the binomial theorem:

Theorem 3: (Binomial Theorem) Let $a$ and $b$ be positive real numbers and $n$ a natural number. Then $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}. $$

Proof of Theorem 2 (using the binomial theorem): Let $n$ be a natural number. Then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} x^n &= \lim_{h\to 0} \frac{(x-h)^n - x^n}{h} \\ &= \lim_{h\to 0} \frac{ \left[ \sum_{k=0}^{n} \binom{n}{k} x^k h^{n-k} \right] - x^n}{ h} && \text{(Binomial Theorem)} \\ &= \lim_{h\to 0} \sum_{k=0}^{n-1} \binom{n}{k} x^k h^{n-k-1} \\ &= \binom{n}{n-1} x^{n-1} && \text{(all other terms have a factor of $h$)} \\ &= n x^{n-1}, \end{align} which is the claimed result. $\square$

I have several complaints about this approach:

1. Many elementary calculus students struggle with even basic arithmetic operations. Throwing binomial coefficients, factorials, and sigma notation at them is at this point in the class is a little unfair, unless one is willing to move the sequences and series material to the beginning of the class (this material is usually taught near the end of the second semester of calculus at American institutions). The sigma notation can be avoided, but it gets replaced with ellipses, which, in my opinion, is less rigorous and a bit hand-wavy.

2. The binomial theorem, itself, requires proof, and the classical proof is by induction:

   Proof of Theorem 3: As a basis for induction, note that $$(a+b)^1 = a+b = \binom{1}{0} a^0 b^1 + \binom{1}{1} a^1 b^0 = \sum_{k=0}^{1} \binom{1}{k} a^k b^{n-k}. $$ Suppose that there is some natural number $k$ such that $$ (a+b)^k = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}. $$ Then \begin{align} (a+b)^{k+1} &= (a+b) (a+b)^k \\ &= (a+b) \sum_{k=0}^{n} \binom{n}{k} a^k b^{k-n} \\ &= \big[\kern-2pt\big[ \text{several tedious steps involving combinatorial identities} \big]\kern-2pt\big] \\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} a^k b^{n+1-k}, \end{align} as desired. $\square$

   Note that I've skipped much of the computational detail here—a lot of those details are, themselves, rather off-topic for a calculus class. In order to use the binomial to prove the product rule, I also have to discuss with students the intermediate results that I've skipped here.

I am not going to argue that proving the product rule in calculus requires induction (I can't think of an alternative proof off the top of my head, but that doesn't mean anything). On the other hand, I think that induction is the best way to obtain the product rule (even if you choose to invoke the binomial theorem, there is an induction hiding behind the scenes), and it provides a good excuse to introduce induction.

In an American "Introduction to Proofs" or elementary "Discrete Mathematics" course, either of which is usually offered after students have taken calculus, I think that the first proof offers a nice call-back to calculus. Students should already know the product rule, and may even have seen some kind of proof (probably a mysterious proof involving the binomial theorem, which they probably won't have seen a proof of). Thus providing this as a first or second example gives some context for induction, and helps to connect earlier mathematical knowledge to new material.

I often teach a short unit on induction in elementary calculus classes—it comes up very naturally when trying to prove a version of the power rule (i.e. the version pertaining to functions of the form $x \mapsto x^n$ for $n\in\mathbb{N}$):

Lemma 1: (Product Rule) Let $f$ and $g$ be real valued functions on $\mathbb{R}$. At any point $x$ where both $f$ and $g$ are differentiable, $$ (fg)'(x) = f'(x) g(x) + f(x) g'(x). $$

As it is not really germane to this discussion, I'll assume that the product rule has already been proved.

Theorem 2: (Power Rule, version 1) For any natural number $n$, $$\frac{\mathrm{d}}{\mathrm{d}x} x^n = nx^{n-1}.$$

Proof of Theorem 2 (by induction): As a basis for induction, note that $$ \frac{\mathrm{d}}{\mathrm{d}x} x^1 = \lim_{h\to 0} \frac{(x+h)^1 - x^1}{h} = \lim_{h\to 0} \frac{h}{h} = 1 = x^0. $$ Suppose that there is some natural number $k$ such that $$ \frac{\mathrm{d}}{\mathrm{d}x} x^k = k x^{k-1}. $$ Then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} x^{k+1} &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ x \cdot x^k \right] \\ &= \left[ \frac{\mathrm{d}}{\mathrm{d}x} x \right] x^k + x \left[ \frac{\mathrm{d}}{\mathrm{d}x} x^k \right] && \text{(Product Rule)} \\ &= 1 \cdot x^k + k x \cdot x^{k-1} && \text{(Induction Hypothesis)} \\ &= (k+1) x^k, \end{align} which is the desired result. $\square$

It is worth noting, perhaps, that most elementary calculus texts prove this version of the power rule by appealing to the binomial theorem:

Theorem 3: (Binomial Theorem) Let $a$ and $b$ be positive real numbers and $n$ a natural number. Then $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}. $$

Proof of Theorem 2 (using the binomial theorem): Let $n$ be a natural number. Then \begin{align} \frac{\mathrm{d}}{\mathrm{d}x} x^n &= \lim_{h\to 0} \frac{(x-h)^n - x^n}{h} \\ &= \lim_{h\to 0} \frac{ \left[ \sum_{k=0}^{n} \binom{n}{k} x^k h^{n-k} \right] - x^n}{ h} && \text{(Binomial Theorem)} \\ &= \lim_{h\to 0} \sum_{k=0}^{n-1} \binom{n}{k} x^k h^{n-k-1} \\ &= \binom{n}{n-1} x^{n-1} && \text{(all other terms have a factor of $h$)} \\ &= n x^{n-1}, \end{align} which is the claimed result. $\square$

I have several complaints about this approach:

1. Many elementary calculus students struggle with even basic arithmetic operations. Throwing binomial coefficients, factorials, and sigma notation at them is at this point in the class is a little unfair, unless one is willing to move the sequences and series material to the beginning of the class (this material is usually taught near the end of the second semester of calculus at American institutions). The sigma notation can be avoided, but it gets replaced with ellipses, which, in my opinion, is less rigorous and a bit hand-wavy.

2. The binomial theorem, itself, requires proof, and the classical proof is by induction:

   Proof of Theorem 3: As a basis for induction, note that $$(a+b)^1 = a+b = \binom{1}{0} a^0 b^1 + \binom{1}{1} a^1 b^0 = \sum_{k=0}^{1} \binom{1}{k} a^k b^{n-k}. $$ Suppose that there is some natural number $k$ such that $$ (a+b)^k = \sum_{k=0}^{n} \binom{n}{k} a^k b^{n-k}. $$ Then \begin{align} (a+b)^{k+1} &= (a+b) (a+b)^k \\ &= (a+b) \sum_{k=0}^{n} \binom{n}{k} a^k b^{k-n} \\ &= \big[\kern-2pt\big[ \text{several tedious steps involving combinatorial identities} \big]\kern-2pt\big] \\ &= \sum_{k=0}^{n+1} \binom{n+1}{k} a^k b^{n+1-k}, \end{align} as desired. $\square$

   Note that I've skipped much of the computational detail here—a lot of those details are, themselves, rather off-topic for a calculus class. In order to use the binomial to prove the product rule, I also have to discuss with students the intermediate results that I've skipped here.

I am not going to argue that proving the power rule in calculus requires induction (I can't think of an alternative proof off the top of my head, but that doesn't mean anything). On the other hand, I think that induction is the best way to obtain the product rule (even if you choose to invoke the binomial theorem, there is an induction hiding behind the scenes), and it provides a good excuse to introduce induction.

In an American "Introduction to Proofs" or elementary "Discrete Mathematics" course, either of which is usually offered after students have taken calculus, I think that the first proof offers a nice call-back to calculus. Students should already know the power rule, and may even have seen some kind of proof (probably a mysterious proof involving the binomial theorem, which they probably won't have seen a proof of). Thus providing this as a first or second example gives some context for induction, and helps to connect earlier mathematical knowledge to new material.