How to prove this identity about log and exp? $$\sum_{k=1}^{m}\sum_{i_1+i_2+\dots+i_k=m}\left(\frac{1}{i_1!}\frac{1}{i_2!}\dots\frac{1}{i_k!}\right)\times\frac{(-1)^{k+1}}{k}=\begin{cases}1,& m=1\\ 0,& m>1\end{cases}$$
where $i_1,\dots,i_k$ are positive integers.
The motivation is to prove that $\log(\exp(x))=x$ in formal power series.
\begin{align*}
x=\log(e^x)=&\sum_{n=1}^\infty (-1)^{n+1}\frac{(e^{x}-1)^n}{n}=\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}(\sum_{m=1}^{\infty}\frac{x^m}{m!})^n                      \\
=&\sum_{m=1}^\infty \sum_{k=1}^{m}(\sum_{i_1+i_2+\dots+i_k=m}\left(\frac{1}{i_1!}\frac{1}{i_2!}\dots\frac{1}{i_k!}\right)\times\frac{(-1)^{k+1}}{k} )x^m
\end{align*}
 A: We seek to show that
$$\sum_{k=1}^{m}\sum_{q_1+q_2+\dots+q_k=m}\left(\frac{1}{q_1!}
\frac{1}{q_2!}\cdots\frac{1}{q_k!}\right)
\times\frac{(-1)^{k+1}}{k}=\begin{cases}1,
& m=1\\ 0,& m>1\end{cases}$$
where $q_1,\dots,q_k$ are positive integers.
Write this as
$$\sum_{k=1}^{m} \frac{(-1)^{k+1}}{k}
\sum_{q_1+q_2+\cdots+q_k=m}
\frac{1}{q_1!}\frac{1}{q_2!}
\cdots\frac{1}{q_k!}.$$
We then get from first principles,
$$[w^m] \sum_{k=1}^m \frac{(-1)^{k+1}}{k}
(\exp(w)-1)^k
\\ = [w^m] \sum_{k=1}^m \frac{(-1)^{k+1}}{k}
\sum_{p=0}^k {k\choose p} \exp(pw) (-1)^{k-p}.$$
We may omit $p=0$ because with $m\ge 1$ there is no contribution to the
coefficient extractor:
$$[w^m] \sum_{k=1}^m \frac{(-1)^{k+1}}{k}
\sum_{p=1}^k {k\choose p} \exp(pw) (-1)^{k-p}
\\ = [w^m] \sum_{p=1}^m (-1)^{p+1} \exp(pw)
\sum_{k=p}^m \frac{1}{k} {k\choose p}
\\ = [w^m] \sum_{p=1}^m \frac{(-1)^{p+1}}{p} \exp(pw)
\sum_{k=p}^m {k-1\choose p-1}.$$
We have for the inner sum
$$\sum_{k=0}^{m-p} {k+p-1\choose p-1}
= [z^{m-p}] \frac{1}{1-z} \sum_{k\ge 0} {k+p-1\choose p-1} z^k
\\ = [z^{m-p}] \frac{1}{1-z} \frac{1}{(1-z)^p}
= [z^{m-p}] \frac{1}{(1-z)^{p+1}}
= {m\choose p}.$$
We thus obtain
$$[w^m] \sum_{p=1}^m \frac{(-1)^{p+1}}{p} \exp(pw) {m\choose p}
= \frac{1}{m} [w^{m-1}]
\sum_{p=1}^m (-1)^{p+1} \exp(pw) {m\choose p}.$$
We get two cases, first case $m=1$, which yields
$$ [w^0] (-1)^2 \exp(w) {1\choose 1} = 1 $$
as desired. For $m\gt 1$, second case, we may restore $p=0$ as we get no contribution to the coefficient extractor in that case:
$$\frac{1}{m} [w^{m-1}]
\sum_{p=0}^m (-1)^{p+1} \exp(pw) {m\choose p}
\\ = - \frac{1}{m} [w^{m-1}] (1-\exp(w))^m = 0$$
because
$$(1-\exp(w))^m = (-1)^m w^m + \cdots.$$
This concludes the argument.
A: Multinomial Theorem : For $i_1+i_2+\dots+i_k=m$, let ${m\choose i_1~\dots~i_k}=\frac{m!}{i_1!\dots i_k!}$ be the multinomial coefficients, then the following generalizes Newton's Binomial's Theorem
\begin{align*}
(x_1+\dots+x_k)^m=\sum_{i_1+\dots+i_k=m} {m\choose i_1~\dots~i_k}\prod_{j=1}^k x_j^{i_j}
\end{align*}
This can be proven by induction on $k$ from the binomial theorem (which can be proven by induction on $m$).
Using this you get
\begin{align*}
&\sum_{k=1}^m \sum_{i_1+\dots+i_k=m} \frac{1}{m!}{m\choose i_1~\dots~i_k}\times \frac{(-1)^{m+1}}{k}\\
=&\sum_{k=1}^m  \frac{1}{m!}\times \frac{(-1)^{m+1}}{k}\sum_{i_1+\dots+i_k=m}{m\choose i_1~\dots~i_k}1^{i_1}\dots 1^{i_k}\\
=&\sum_{k=1}^m  \frac{1}{m!}\times \frac{(-1)^{m+1}}{k}k^m\\
=&\frac{(-1)^{m+1}}{m!}\sum_{k=1}^m k^{m-1}
\end{align*}
As mentionned by @ancientmathematician, this will not lead to what your want but shows you a way to use the multinomial Theorem for such cases.
A: The easiest way to prove $\log \exp x = x$ as an identity of formal power series is to take derivatives. Observe that we have $\frac{d}{dx} \log \left( 1 + x \right) = \frac{1}{1 + x}$ (as an identity of formal power series) and $\frac{d}{dx} \exp(x) = \exp(x)$; the formal chain rule then gives
$$\frac{d}{dx} \log \exp x = \frac{1}{\exp(x)} \exp(x) = 1$$
from which it follows that $\log \exp x = x$ after plugging in $x = 0$ to verify that the constant term is correct. In the middle technically we are writing the above expression as $\log (1 + (\exp x - 1))$ to keep everything in the realm of formal power series.
