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. 

You want to prove that
$$\sum_{k=1}^n(k^2+1)k! = n(n+1)!.$$
First case: if $n = 1$, then $(1^2+1)1! = 2 = 1 \cdot 2 = n(n+1)!$.
Then by induction, if the formula holds for $n-1$ you find:
$$\sum_{k=1}^n(k^2+1)k! = \sum_{k=1}^{n-1}(k^2+1)k! + (n^2+1)n! = (n-1)(n-1+1)! + (n^2+1)n! = (n-1)n! + (n^2+1)n! = (n+n^2)n! = n(n+1)n! = n(n+1)!$$
