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. 
 A: 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. 
A: 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:
$$\sum_{k=1}^n(k^2+1)k! = \sum_{k=1}^{n-1}(k^2+1)k! + (n^2+1)n! = (n-1)(n-1+1)! + (n^2+1)n! = (n-1)n! + (n^2+1)n! = (n+n^2)n! = n(n+1)n! = n(n+1)!$$
A: Alternatively:
$$2 · 1! + 5 · 2! + 10 · 3! + . . . + (n
^2 + 1)n! =\sum_{i=1}^n(i^2+1)i!=\\
\sum_{i=1}^n[(i+1)^2-i-i]i!=\\
\color{red}{\sum_{i=1}^n(i+1)\cdot (i+1)!-\sum_{i=1}^n i\cdot i!}-\color{blue}{\sum_{i=1}^n (i+1-1)\cdot i!}=\\
\color{red}{(n+1)(n+1)!-1}-\color{blue}{(n+1)!+1}= n(n + 1)!$$
