## [ Sums of numerical sequences and series of differences] Products and factorials

Find the sum $1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!$.
$S=1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!=(2!-1!)+(3!-2!)+\ldots+(n!-(n-1)!)+((n+1)!-n!)=(n+1)!-1!$.

## Answer

$(n+1)!-1$

## Problem