<s>[INST] Proof for maths induction prove: $$ \sum_{r=1}^{n}[r^{2}+1](r!)=n[(n+1)!]$$ for all $n \in N$
prove $n=1$, $(1^2+1)(1!)$ = $1[(1+1)!]$
assume true for $n=k$, $(k^2+1)$$(k!)$= $k$$[(k+1)!]$
I got to here : $[k^{2}+1](k!)+[(k+1)^2+1](k+1)! = (k+1)[(k+2)!]$
 [/INST] If
$$ \sum_{r=0}^{n}(r^{2}+1)r!=n(n+1)!$$
then
$$ \sum_{r=0}^{n+1}(r^{2}+1)r!=\sum_{r=0}^{n}(r^{2}+1)r!+((n+1)^2+1)(n+1)!=$$
$$=n(n+1)!+((n+1)^2+1)(n+1)!=$$
$$=(n+1)!(n+((n+1)^2+1))=$$
$$=(n+1)!(n+n^2+2n+1+1)=$$
$$=(n+1)!(n^2+3n+2)=$$
$$=(n+1)!(n+2)(n+1)=$$
$$=(n+1)(n+2)!$$
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