Prove that 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1, for n >= 1

Here's what I have:

Base case: n = 1.

s(n) = 1(1!) + 2(2!) + ... + n(n!) = (n+1)! - 1

Therefore, (1+1)! - 1 = 1

Correct...

Inductive step:

For n+1,

$\displaystyle n(n!) = (n+1)*(n+1)!$

$\displaystyle s(n+1) = s(n) + (n+1)*(n+1)! $

or...

$\displaystyle s(n+1) = (n+1)! - 1 + (n+1)*(n+1)! $

This is where I am stuck. What next?