$\sum_{n=1}^m n \cdot n$ ! is equal to

The given sum is $S = \sum_{n=1}^{m} n \cdot n!$.

Observe that the general term can be written as $n \cdot n! = (n+1-1) \cdot n! = (n+1) \cdot n! - n!$.

Since $(n+1) \cdot n! = (n+1)!$, the term becomes $(n+1)! - n!$.

The summation is a telescoping series:

$S = \sum_{n=1}^{m} ((n+1)! - n!)$

$S = (2! - 1!) + (3! - 2!) + (4! - 3!) + \dots + ((m+1)! - m!)$

All intermediate terms cancel out, leaving only the last positive term and the first negative term:

$S = (m+1)! - 1!$

$S = (m+1)! - 1$.

Answer: $(m+1) !-1$