An algebra problem by Ραμών Αδάλια

$\begin{aligned}\pi (n)&=\sum _{ j=1 }^{ n }{ \sum _{ i=1 }^{ j }{ i!({ i }^{ 2 }+i+1) } }\\&=\sum _{ j=1 }^{ n }{ \sum _{ i=1 }^{ j }{ i!((i+1)^2-i) }} \\&=\sum _{ j=1 }^{ n }{ \sum _{ i=1 }^{ j }{( i+1)(i+1)!-i (i!)}}&\small\color{#3D99F6}\text{We can see that the sum is telescoping}\\&=\sum _{ j=1 }^{ n }{(( j+1)(j+1)!-1 )}\\&=\left(\sum _{ j=1 }^{ n }{( j+1)!(j+2-1)}\right)-n\\&=\left(\sum _{ j=1 }^{ n }{( j+2)!-(j+1)!}\right)-n&\small\color{#3D99F6}\text{Once again, the sum is telescoping}\\\pi(n)&=((n+2)!-2)-n=(n+2)!-(n+2)\\\pi(7)&=(9!-9)=362871\\&\small\color{#3D99F6}\text{The last 2 digits of }\pi(7)=\boxed{71}\end{aligned}$