Prove $1(1!)+\dots+n(n!) = (n+1)!-1$ using induction

Question

So I'm trying to prove this statement (through induction):

$$1(1!)+2(2!)+\dots +n(n!)=(n+1)!-1$$

But I'm confused with the inductive step here:

$$(n+1)!-1+[(n+1)(n+1)!] = (n+2)!-1$$

What do I do at this point? The factorial is throwing me off, because I can't multiply $(n+1)(n+1)!$ together to simplify the left side, can I?

Answer 1

Hint:

$$(n+1)!+(n+1)(n+1)!=(n+1+1)[(n+1)!]=(n+2)(n+1)!=(n+2)!$$

Answer 2

From the induction step we get: $$(n+1)!-1+[(n+1)(n+1)!]=(n+2)!-1$$ Now let's add and subtract (n+1)! from the right side of the equation, in order to get the form of the induction assumption on the right side too: $$(n+1)!-1+[(n+1)(n+1)!]=(n+2)!+(n+1)!-(n+1)!-1=(n+1)!-1+[(n+2)!-(n+1)!]$$ from the induction assumption we can simplify the equation: $$[(n+1)(n+1)!]=[(n+2)!-(n+1)!]$$ $$[(n+1)(n+1)!]=[(n+2)(n+1)!-(n+1)!]$$ $$[(n+1)(n+1)!]=[(n+2-1)(n+1)!]$$ $$[(n+1)(n+1)!]=[(n+1)(n+1)!]$$

And that's what we were looking for.