Proof that $\dfrac{1}{e^x}=e^{-x}$ without converting it to $e^{x}e^{-x}=1$.

Question

I want to show that $\dfrac{1}{e^x} = e^{-x}$ from the Taylor expansion of $e^x$.

To express $\dfrac{1}{e^x}$ as a power series, I let: $$ \left(\dfrac{1}{0!}x^0 + \dfrac{1}{1!}x^1 + \dfrac{1}{2!}x^2 + \dfrac{1}{3!}x^3 + \cdots \right) \left(b_0 x^0 + b_1 x^1 + b_2 x^2 + b_3 x^3 + \cdots \right) = 1x^0 $$

and start comparing coefficients from $x^0$ upwards. As expected, a pattern starts to emerge that $b_k = \dfrac{(-1)^k}{k!}$, but it gets quite messy quite quickly, so I'm not sure how I can formally prove that this pattern continues for all $k$.

I've thought about using induction, of course, but I can't come up with the correct way to set it up.

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Edit: As @CameronWilliams correctly pointed out below, it would be easier to show that $e^{-x}e^{x}=1$, but I would personally find it more elegant to compute the reciprocal and then recognize that it happens to be $e^{-x}$.

Answer 1

Hint: $$\sum_{n=0}^{\infty}\frac{1}{n!}x^n\cdot\sum_{m=0}^{\infty}\frac{(-1)^m}{m!}x^m = \sum_{k=0}^\infty c_k x^k$$

where

$$c_k = \sum_{n+m=k}\frac{1}{n!}\frac{(-1)^{m}}{m!} = \frac{(-1)^k}{k!}\sum_{n=0}^{k}{k\choose n}(-1)^{n}$$

The last sum can be evaluated using the Binomial formula.

Answer 2

This isn't quite what you're looking for, but it may be interesting to you and was too long to put in a comment. Rather than by an explicit series, it's also possible to define $y(x) = e^x$ as the unique solution (say, defined on the real line) of $y = y'$ with $y(0) = 1$. For any $a$, the function $y_a(x) = \frac{1}{y(a)}\, y(x + a)$ then also clearly satisfies $y_a'(x) = y_a(x)$ and $y_a(0) = 1$. Hence $y_a(x) = y(x)$; that is, $e^{x+a} = e^a e^x$ for fixed $a$. Taking $a = -x$ for some fixed $x$ gives the required result.