TITLE: Proof of a combinatorial equation
QUESTION [8 upvotes]: How can we use elementary methods to prove that
$$\sum_{i = 2}^{n}{{n \choose i} i! n^{n - i}} = \sum_{i = 1}^{n - 1}{{n \choose i}i^i (n - i)^{n - i}}$$
for any integer $n \geq 0$?
The values of each side for fixed $n$ are 0, 0, 2, 24, 312, 4720, ... (A001864 - OEIS).

REPLY [2 votes]: Here is an alternative.
Define the functions $A_n(t)=\sum_k\binom{n}kt(t+k)^{k-1}(n-k)^{n-k}, B_n(t)=\sum_k\frac{n!}{(n-k)!}(t+n)^{n-k}$ and
$C_n(t)=\sum_k\binom{n}k(t+k)^k(n-k)^{n-k}$. 
From $(t+k)^k=t(t+k)^{k-1}+k(t+k)^{k-1}$ and on the basis of Abel's identity, 
one gets $C_n(t)=A_n(t)+nC_{n-1}(t+1)=(t+n)^n+nC_{n-1}(t+1)$. It is easy to check that $B_n(t)=(t+n)^n+nB_{n-1}(t+1)$. Since both $B_n$
and $C_n$ satisfy the same initial conditions, it follows $B_n(t)=C_n(t)$. So, $B_n(0)=C_n(0)$ gives the desired result.