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)*.
A simpler proof can be found in the solution of Exercise 1 **(a)** in
UMN Fall 2018 Math 5705 midterm #3
(this one was found by Tomoya Imaizumi, a student in that class).
(My exercise uses two rational numbers $x$ and $y$ instead of the
indeterminates $X$ and $Y$, but the proof does not care.)
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*.
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.