How can we 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).
How can we 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).