Show that $2 · 1! + 5 · 2! + 10 · 3! + . . . + (n ^2 + 1)n! = n(n + 1)!$ Show  that
$$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! = n(n + 1)!$$
I got stuck after the n+1 part.
$$((n+1)^2 + 1)(n+1)! = (n+1)(n+2)!.$$
I'm not sure how to proceed from this part. 

proof by Induction $$P(1):=2\cdot1!=1\cdot(1+1)!$$
Now we assume it is true for $P(n)$ then we proof it is true for $P(n+1)$, $$P(n+1):=2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! + ((n+1)^2+1)(n+1)! = (n+1)((n + 1)+1)!$$
Use the hypothesis of induction $P(n)$ and LHS of above expression reduces to $$n(n+1)!+((n+1)^2+1)(n+1)! = (n+1)!(n+n^2+2n+2) \\ =(n+1)!(n^2+3n+2)=(n+1)!(n+2)(n+1) \\ =(n+1)(n+2)! $$
Hence we get the RHS of the $P(n+1)$. So we have proved $P(n+1)$ is true whenever $P(n)$ is true. 
