TITLE: How to prove that $\exp(x)$ and $\log(x)$ are inverse?
QUESTION [8 upvotes]: How does one prove that the exponential and logarithmic functions are inverse using the definitions:
$$e^x= \sum_{i=0}^{\infty} \frac{x^i}{i!}$$
and 
$$\log(x)=\int_{1}^{x}\frac{1}{t}dt$$
My naive approach (sort of ignoring issues of convergence) is to just apply the definitions straightforwardly, so in one direction I get:
\begin{align}\log(e^x)&=\int_{1}^{e^x}\frac{1}{t}dt\\
&=\int_{0}^{e^x-1}\frac{1}{1+t}dt\\
&=\int_0^{e^x-1}\sum_{j=0}^\infty (-1)^jt^jdt \\
&=\sum_{j=0}^\infty (-1)^j \int_0^{e^x-1} t^jdt\\
&=\sum_{j=0}^\infty \frac{(-1)^j}{j+1}(e^x-1)^{j+1}\\
&=\sum_{j=0}^\infty \frac{(-1)^j}{j+1} \sum_{k=0}^{j+1} \frac{n!}{k!(n-k)!}e^{x(n-k)}(-1)^k\\
&=\sum_{j=0}^\infty \sum_{k=0}^{j+1}  \frac{(-1)^{j+k}n!}{(j+1)k!(n-k)!} e^{x(n-k)}\\
&=\sum_{j=0}^\infty \sum_{k=0}^{j+1}  \frac{(-1)^{j+k}n!}{(j+1)k!(n-k)!} \sum_{\ell=0}^\infty \frac{(-1)^\ell}{\ell !}(n-k)^\ell x^\ell\\
&=\sum_{j=0}^\infty \sum_{k=0}^{j+1} \sum_{\ell=0}^\infty \frac{(-1)^{j+k+\ell}n!(n-k)^\ell x^\ell}{(j+1)k!\ell!(n-k)!} \end{align}
and I cant see at all that this is equal to $x$. My guess is I'm going about this all wrong.

REPLY [2 votes]: We use the fact that $g(x)=\exp(x)$ is the unique function $g:\mathbb{R} \rightarrow (0,\infty)$ such that $g'(x)=g(x)$ and $g(0)=1.$
Since $f(x)=\log(x)$ is evidently bijective, it must have the inverse $f^{-1}:\mathbb{R} \rightarrow (0,\infty)$. By the well-known theorem on the derivative of the inverse function, $$(f^{-1}){'}(x)=\frac{1}{f'(f^{-1}(x))}=\frac{1}{1/(f^{-1}(x))}=f^{-1}(x)$$ for every $x>0$. Furthermore, since $f(1)=\log(1)=0,$ $f^{-1}(0)= 1$. 
It follows that $f^{-1}(x)$ must be $\exp(x).$