How can I prove that $\ln(e^x)=x$ using the Taylor series of $e^x$ and $\ln(1+x)$? I'm stuck to prove that $\ln(e^x)=x$ using the facts that
\begin{align}
e^x &= \sum_{k=0}^\infty \frac{x^k}{k!} \quad\text{and}  \\
\ln(1+x) &= \sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}x^k.
\end{align}
I tried as follows:
$$\ln(e^x)=\ln\left(\sum_{k=0}^\infty \frac{x^k}{k!}\right)=\ln\left(1+\sum_{k=1}^\infty \frac{x^k}{k!}\right)=\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k}\left(\sum_{n=1}^\infty \frac{x^n}{n!}\right)^k$$
but the last sum looks very complicate to simplify. Any ideas?
 A: Near $x = 0$, the sum expands as
\begin{align*}
\log(e^x)
&= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n_1, \cdots, n_k = 1}^{\infty} \frac{x^{n_1+\cdots+n_k}}{n_!! \cdots n_k!} \\
&= \sum_{l=1}^{\infty} \Bigg( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{\substack{n_1 \geq 1, \cdots, n_k \geq 1 \\ n_1 + \cdots + n_k = l}} \frac{1}{n_1! \cdots n_k!} \Bigg) x^l \\
&= \sum_{l=1}^{\infty} \Bigg( \sum_{k=1}^{l} \frac{(-1)^{k-1}}{k} \sum_{\substack{n_1 \geq 1, \cdots, n_k \geq 1 \\ n_1 + \cdots + n_k = l}} \binom{l}{n_1,\cdots,n_k} \Bigg) \frac{x^l}{l!}
\end{align*}
Except for the obvious algebraic manipulation, we remark that the upper limit of the sum for $k$ is replaced by $l$ in the last line, simply because the summand is zero when $k > l$.
We focus on the innermost sum. This counts $k!$ the number of ways of partitioning the set $\{1,\cdots,l\}$ into $k$ non-empty unordered sets, which is exactly the Stirling number of the second kind $\big\{ {l \atop k} \big\}$, and so, the problem reduces to showing that $ \sum_{k=1}^{\infty} (-1)^{k-1} (k-1)! \big\{ {l \atop k} \big\} = \delta_{1l}$. To avoid possible circular argument, however, we will not utilize this observation.
Returning to computing the innermost sum, we note that this is a reminiscence of the sum appearing in the multinomial expansion, although it is not entirely the same. Indeed, this sum is missing terms corresponding to the indices $(n_1, \cdots, n_k)$ with at least one of them being zero. For a precise relationship between this sum and the multinomial expansion, we observe that
\begin{align*}
k^l
= (\overbrace{1 + \cdots + 1}^{k\text{ terms}})^l
&= \sum_{\substack{n_1 \geq 0, \cdots, n_k \geq 0 \\ n_1 + \cdots + n_k = l}} \binom{l}{n_1,\cdots,n_k} \\
&= \sum_{j=0}^{k} \binom{k}{j} \sum_{\substack{n_1 \geq 1, \cdots, n_j \geq 1 \\ n_1 + \cdots + n_j = l}} \binom{l}{n_1,\cdots,n_j}.
\end{align*}
In light of the inverse binomial transform, this yields
$$ \sum_{\substack{n_1 \geq 1, \cdots, n_k \geq 1 \\ n_1 + \cdots + n_k = l}} \binom{l}{n_1,\cdots,n_k} = \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^l. \tag{*} $$
Plugging this back, the problem boils down to establishing the identity
$$ \sum_{k=1}^{l} \frac{(-1)^{k-1}}{k} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^l \quad \stackrel{?}{=} \quad \begin{cases} 1, & l = 1; \\ 0, & l \geq 2. \end{cases} \tag{$\diamond$} $$
To compute this double sum, we interchange the order of summation. Then
\begin{align*}
\sum_{k=1}^{l} \frac{(-1)^{k-1}}{k} \sum_{j=0}^{k} (-1)^{k-j} \binom{k}{j} j^l
&= \sum_{j=1}^{l} (-1)^{j-1} j^l \sum_{k = 1}^{l} \frac{1}{k} \binom{k}{j} \\
&= \sum_{j=1}^{l} (-1)^{j-1} j^{l-1} \sum_{k = 1}^{l} \binom{k-1}{j-1} \\
&= \sum_{j=1}^{l} (-1)^{j-1} j^{l-1} \binom{l}{j} \tag{$\star$}
\end{align*}
where the last line follows from the hockey-stick identity for the binomial coefficients. However, in the case $l \geq 2$, we recognize $(\star)$ as a special case of $\text{(*)}$, corresponding to $l$ indices $n_1, \cdots, n_l \geq 1$ summing to $l-1$. Since there is no such $l$-tuple $(n_1, \cdots, n_l)$, the corresponding sum evaluates to $0$ as required. When $l = 1$, manual computation of $(\star)$ gives $1$. Therefore $(\diamond)$ is established, proving the desired identity. $\square$
