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.