Discrete Mathematics induction proof: $1! + 2! + 3! + ldots + n! < (n + 1)!$

I am trying to solve a induction proof, and I got stuck at the end. Some help would be great. This is the question, and what I did so far:

Statement: For all integers $n \geq 1$ we have $1!+2!+3!+ \ldots +n!<(n+1)!$.

Proof: Induction over $n$. Introduce the name $P(n)$ for the statement $1!+2!+3!+ \ldots +n!<(n+1)!$. We shall prove, by mathematical induction, that $\forall n \geq 1: P(n)$.

$P(n)$ and test with $1$ so that $P(1): 1!<(1+1)! \implies 1<2$ (true)

Assume $P(k)$ so that we have $P(k): 1!+2!+3!+ \ldots +k!<(k+1)!$, and we want to prove $P(k+1):1!+2!+3!+....+(k+1)!<(k+2)!$. It is here where I don't know how to continue to prove the statement.

Please help

Answer

Hint: by induction hypothesis $1!+2!+...+n!<(n+1)!$

Now $1!+2!+...+(n+1)!<2(n+1)!<(n+2)!$