TITLE: Exponential and log functions compose to identity
QUESTION [5 upvotes]: How to prove that the exponential function and the logarithm function are the inverses of each other? I want it the following way. We must use the definition as power series, and must verify that all the terms of the composition except the coefficent of $z$ vanish, and that the first degree term is $1$.
I can write down the proof for the coefficient of $x^n$ for arbitrary but fixed $n$ by explicit verification. But how to settle this for all $n$ at one go?

REPLY [2 votes]: Qiaochu Yuan's proof is without doubt the most elegant. However, here is a purely formal argument using the derivative:
Let $L(x) = \log \frac 1{1-x} = \sum_{n\geq 1} \frac{x^n}n$. I will show that $\exp(L(x))=\frac 1{1-x}$.
Let $u(x)=\exp(L(x))$. By the chain rule for differentiation, we have
$$
u'(x) = L'(x)u(x).
$$
Noting that $L'(x)=(1-x)^{-1}$, the above can be rewritten as
$$
u(x)=(1-x)u'(x).
$$
Suppose $u(x)=\sum_{n\geq 0} a_nx^n$, then comparing the coefficients of each power of $x$ in the preceding identity gives
$$
a_n = (n+1)a_{n+1} - na_n
$$ for each $n\geq 0$.
Also, $a_0=1$.
It follows that $a_n=1$ for all $n$, so $u(x)=\frac 1{1-x}$, as desired.
The reverse identity, $\log(\exp(x))=x$ is simpler; the chain rule tells us that the derivative of $\log(\exp(x))=1$.