combinatorics - Sum identity using Stirling numbers of the second kind

Experimenting with series representations of $e^{x e^x}$ I came across the two seemingly different power series
$$e^{x e^x} = \sum_{n=0}^{\infty} x^n \sum_{k=0}^{n} \frac{(n-k)^k}{(n-k)! \cdot k!}$$
and
$$e^{x e^x} = \sum_{n=0}^{\infty} x^n \sum_{k=0}^{n} \frac{1}{(n-k)!} \sum_{i=0}^{k} \frac{1}{(k-i)!} {k-i \brace i}\,.$$
They should be and are in fact identical. By equating the coefficients it is obvious that
$$\sum_{k=0}^{n} \frac{(n-k)^k}{(n-k)! \cdot k!} = \sum_{k=0}^{n} \frac{1}{(n-k)!} \sum_{i=0}^{k} \frac{1}{(k-i)!} {k-i \brace i}\,.$$
Going through some values of $n$ reassures the correctness of the equation, however I have no idea how one could show this algebraically. An equivalent equation would be
$$\sum_{k=0}^{n} {n \choose k} (n-k)^k = \sum_{k=0}^{n} {n \choose k} \sum_{i=0}^{k} {k \choose i} \sum_{j=1}^{i} (-1)^{i-j} {i \choose j} j^{k-i}\,,$$
which is obtained through multiplication of both sides by $n!$ and an explicit formula for the Stirling numbers. No matter what I try, I seem to run into a brick wall. How can I (algebraically) show that the equation is correct for arbitrary $n$?

Answer

Here is another variation of the theme. In order to prove the binomial identity
\begin{align*}
\sum_{k=0}^{n} \binom{n}{k} (n-k)^k
=\sum_{k=0}^{n} \binom{n}{k} \sum_{i=0}^{k}\binom{k}{i}
\sum_{j=0}^{i}\binom{i}{j} (-1)^{i-j} j^{k-i}\tag{1}
\end{align*}

we take a look at a binomial inverse pair. Let $F(x)=\sum_{n=0}^\infty f_n \frac{x^n}{n!}$ and $G(x)=\sum_{n=0}^\infty g_n \frac{x^n}{n!}$ be exponential generating functions. We consider the following

Binomial inverse pair:
\begin{align*}
F(x)&=G(x)e^x&G(x)&=F(x)e^{-x}\\
f_n&=\sum_{k=0}^{n}\binom{n}{k}g_k&g_n&=\sum_{k=0}^{n}\binom{n}{k}(-1)^{n-k}f_k\qquad \tag{2}
\end{align*}

This simple one and many other examples of binomial inverse pairs can be found e.g. in the classic Combinatorial Identities by John Riordan ($1968$).

We obtain from (2) by successively substituting $f_n$ resp. $g_n$
\begin{align*}
f_n&=\sum_{k=0}^{n}\binom{n}{k}g_k\tag{3}\\
&=\sum_{k=0}^{n}\binom{n}{k}\sum_{i=0}^k\binom{k}{i}(-1)^{k-i}f_i\\
&=\sum_{k=0}^{n}\binom{n}{k}\sum_{i=0}^k\binom{k}{i}(-1)^{k-i}\sum_{j=0}^{i}\binom{i}{j}g_j\tag{4}
\end{align*}

Setting $g_k=k^{n-k}$ we obtain from (3) and (4)

The following is valid for $n\geq 0$:
\begin{align*}
\sum_{k=0}^{n}\binom{n}{k}k^{n-k}=\sum_{k=0}^{n}\binom{n}{k}\sum_{i=0}^k\binom{k}{i}(-1)^{k-i}\sum_{j=0}^{i}\binom{i}{j}j^{i-j}\tag{5}
\end{align*}

Observe the LHS of (5) corresponds to the LHS of (1) by replacing $k\rightarrow n-k$. So, we are nearly finished.

Finally we have to show the RHS of (5) is equal to the RHS of (1). Be careful, although they look similarly, they are not the same!

The validity of
\begin{align*}
\sum_{k=0}^{n} \binom{n}{k} \sum_{i=0}^{k}\binom{k}{i}\sum_{j=0}^{i}\binom{i}{j} (-1)^{i-j} j^{k-i}
=\sum_{k=0}^{n}\binom{n}{k}\sum_{i=0}^k\binom{k}{i}\sum_{j=0}^{i}\binom{i}{j}(-1)^{k-i}j^{i-j}
\end{align*}
is shown in this question and the claim follows.

Note: During analysis of the identity, when I had the association with binomial inverse pairs I was pretty sure, that the expressions (4) and (5) should already result in the RHS of (1). It was really amazing to see that this was not the case and it needed a considerable amount of effort to complete the answer.