Prove that $\exp(\log(\frac{1}{1-x})) = \frac{1}{1-x}$ I am trying to prove this directly by comparing the coefficients in the two series rather than using formal calculus.
Here is what I have so far, but I think I made a mistake:
\begin{align*}
e^{\log\frac{1}{1-x}} &= \sum_{n\geq0}\frac{1}{n!}\left(\log\frac{1}{1-x}\right)^n\\
&= 1+\sum_{n\ge1}\frac{1}{n!}\sum_{k\geq 1}\sum_{i_1+\dotsb+i_n=k}\frac{x^k}{i_1\dotsb i_n}
\end{align*}
Is this correct so far? If so, how do I proceed?
EDIT: \begin{align*}
\log\left(\frac{1}{1-x}\right) &= \sum_{k\geq1}\frac{x^k}{k}\\
e^x &= \sum_{n\geq 0} \frac{x^n}{n!}
\end{align*}
The composition is well defined since the constant term of $\log(\frac{1}{1-x})$ is 0.
 A: A combinatorial proof.
Write $$-\log(1-x)=\sum_{n=1}^{\infty}\frac{(n-1)!}{n!}x^n.$$
The $(n-1)!$ is the number of ways to seat people in a circle of size $n,$ where rotations are considered the same.
So:
$$(-1)^k\log^k(1-x)=\sum_{n=0}^\infty\frac{a_{k,n}}{n!}x^n$$
Where $a_{k,n}$ is the number of ways to seat $n$ people around $k$ ordered circles, no table empty. You have $a_{0,0}=1,$ and $a_{k,n}=0$ for $k<n.$
Then $$\begin{align}\exp(-\log(1-x))&=\sum_{k=0}^{\infty}\frac{1}{k!}\sum_{n=k}\frac{a_{k,n}}{n!}x^n\\
&=\sum_{n=0}^{\infty}\frac{x^n}{n!}\sum_{k=0}^n \frac{a_{k,n}}{k!}
\end{align} $$
So you need to prove:
$$\sum_{k=0}^n \frac{a_{k,n}}{k!}=n!\tag1$$ for all $n.$
Once you have that, you get:
$$\exp(-\log(1-x))=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}.$$
Now, $\frac{a_{k,n}}{k!}$ is the number of ways to arrange $n$ people around $k$ unordered circles.
So we want to find a way to map the set of all permutations of $\{1,2,\dots,n\}$ to the set of all ways to place $n$ people in an unordered set of circles, with any number of circles.
But when we write a permutation as cycles, that is just what we are doing. When $n=6,$ for example, the permutation $(123)(45)(6)$ can be thought of as putting 123 in that order, clockwise, around one circle, $45$ puts people $45$ in one circle, and $6$ sits alone at a circle.

The key is the combinatorial treatment of exponential generating functions, and what $f^n(x)$ means when $f(x)$ is a combinatorial generating function.
We can prove the combinatorial definition of $a_{k,n}$ by induction on $k.$ You just need the property that:
$$\sum_{n=0}^\infty \frac{b_n}{n!}x^n\sum_{m=0}^\infty \frac{c_m}{m!}x^m=\sum_{n=0}^{\infty} \frac{x^n}{n!} \sum_{j=0}^n\binom nk b_jc_{n-j}$$
Then if $c_{n}=a_{k,n}$ and $b_n=(n-1)!$ (and $b_0=0.$) Then you get:
$$a_{k+1,n} =\sum_{j=0}^n\binom nj b_ja_{k,n-j}\tag2$$
It is easy to see if $a_{k,n}$ is the number of ways to put $n$ people in $k$ ordered circles, the the sum on the right adds, for each $j,$ the number of ways of putting $j$ people around a first circle, times the number of ways of putting the remains people in $k$ circles.

There might be a non-combinatorial proof of $(1)$ directly from the recurrence $(2).$ I don't see it immediate, but it seems like some induction might work.
