Let $\mathbb{N}=\left\{  0,1,2,\ldots\right\}  $. The following fact I have
seen referred to as the "Cauchy identity":

> **Theorem 1.** Let $n\in\mathbb{N}$. Then,
\begin{equation}
\sum_{k=0}^{n}\dbinom{n}{k}\left(  X+k\right)  ^{k}\left(  Y-k\right)
^{n-k}=\sum_{t=0}^{n}\dfrac{n!}{t!}\left(  X+Y\right)  ^{t}
\end{equation}
in the polynomial ring $\mathbb{Z}\left[  X,Y\right]  $.

One proof of Theorem 1 can be found in [Darij Grinberg, *6th QEDMO 2009,
Problem 4 (the Cauchy identity)*](http://www.cip.ifi.lmu.de/~grinberg/QEDMO6P4long.pdf). Alternatively, Theorem 1 is the particular case
(for $\mathbb{L}=\mathbb{Z}\left[  X,Y\right]  $, $S=\left\{  1,2,\ldots
,n\right\}  $ and $x_{s}=1$) of Theorem 2.2 in [Darij Grinberg,
*Noncommutative Abel-like identities*](http://www.cip.ifi.lmu.de/~grinberg/algebra/ncabel.pdf). More directly, it is the particular case (for
$Z=1$) of equality (1) in the latter reference, where I also cite other sources.

> **Corollary 2.** Let $n\in\mathbb{N}$. Then,
\begin{equation}
\sum_{i=0}^{n}\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}=\sum_{i=0}
^{n}\dbinom{n}{i}i!n^{n-i}.
\end{equation}


*Proof of Corollary 2.* Theorem 1 is an equality between two polynomials.
Renaming the summation index $k$ as $i$ in this equality, we obtain
\begin{equation}
\sum_{i=0}^{n}\dbinom{n}{i}\left(  X+i\right)  ^{i}\left(  Y-i\right)
^{n-i}=\sum_{t=0}^{n}\dfrac{n!}{t!}\left(  X+Y\right)  ^{t}
\end{equation}
Substituting $0$ and $n$ for $X$ and $Y$ in this equality, we find
\begin{align*}
& \sum_{i=0}^{n}\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}\\
& =\sum_{t=0}^{n}\dfrac{n!}{t!}n^{t}\\
& =\sum_{i=0}^{n}\underbrace{\dfrac{n!}{\left(  n-i\right)  !}}_{=\dbinom
{n}{i}i!}n^{n-i}\qquad\left(
\begin{array}
[c]{c}
\text{here, we have substituted }n-i\text{ for }t\\
\text{in the sum}
\end{array}
\right)  \\
& =\sum_{i=0}^{n}\dbinom{n}{i}i!n^{n-i}.
\end{align*}
This proves Corollary 2. $\blacksquare$

Are there combinatorial proofs of Corollary 2? I'm pretty sure that the answer
is "Yes", and I suspect that they involve counting some sort of functions from
$\left\{  1,2,\ldots,n\right\}  $ to $\left\{  1,2,\ldots,n\right\}  $ with
some specific conditions on their recurrent values.

>**Corollary 3.** Let $n\in\mathbb{N}$. Then,
\begin{equation}
\sum_{i=2}^{n}\dbinom{n}{i}i!n^{n-i}=\sum_{i=1}^{n-1}\dbinom{n}{i}i^{i}\left(
n-i\right)  ^{n-i}.
\end{equation}


*Proof of Corollary 3.* If $n\leq1$, then both sides are $0$, whence the
equality follows. Hence, we WLOG assume that $n>1$. Thus,
\begin{align*}
& \sum_{i=0}^{n}\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}\\
& =\underbrace{\dbinom{n}{0}}_{=1}\underbrace{0^{0}}_{=1}\underbrace{\left(
n-0\right)  ^{n-0}}_{=n^{n}}+\sum_{i=1}^{n-1}\dbinom{n}{i}i^{i}\left(
n-i\right)  ^{n-i}+\underbrace{\dbinom{n}{n}}_{=1}n^{n}\underbrace{\left(
n-n\right)  ^{n-n}}_{=0^{0}=1}\\
& =n^{n}+\sum_{i=1}^{n-1}\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}+n^{n}.
\end{align*}
Comparing this with
\begin{align*}
& \sum_{i=0}^{n}\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}\\
& =\sum_{i=0}^{n}\dbinom{n}{i}i!n^{n-i}\qquad\left(  \text{by Corollary
2}\right)  \\
& =\underbrace{\dbinom{n}{0}}_{=1}\underbrace{0!}_{=1}\underbrace{n^{n-0}
}_{=n^{n}}+\underbrace{\dbinom{n}{1}}_{=n}\underbrace{1!}_{=1}n^{n-1}
+\sum_{i=2}^{n}\dbinom{n}{i}i!n^{n-i}\\
& =n^{n}+\underbrace{nn^{n-1}}_{=n^{n}}+\sum_{i=2}^{n}\dbinom{n}{i}
i!n^{n-i}=n^{n}+n^{n}+\sum_{i=2}^{n}\dbinom{n}{i}i!n^{n-i},
\end{align*}
we obtain
\begin{align*}
n^{n}+n^{n}+\sum_{i=2}^{n}\dbinom{n}{i}i!n^{n-i}=n^{n}+\sum_{i=1}^{n-1}
\dbinom{n}{i}i^{i}\left(  n-i\right)  ^{n-i}+n^{n}.
\end{align*}
Subtracting $n^{n}+n^{n}$ from both sides of this equality, we obtain
\begin{equation}
\sum_{i=2}^{n}\dbinom{n}{i}i!n^{n-i}=\sum_{i=1}^{n-1}\dbinom{n}{i}i^{i}\left(
n-i\right)  ^{n-i}.
\end{equation}
This proves Corollary 3. $\blacksquare$

Corollary 3 is your claim.