$\begingroup$

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.