Question

Mathematics Inclined Week #1: Difficult Category

Answer 1

Let $n=1$. Then the statement says

$1(1!)=1$
$(1+1)!-1=2!-1=2-1=1$

so it holds for the base case.

Assume it holds for $n=k$, i.e. that

$1(1!)+2(2!)+3(3!)+\cdots+k(k!)=(k+1)!-1$

and use this to show it holds for $n=k+1$. You have

$1(1!)+2(2!)+\cdots+k(k!)+(k+1)(k+1)!=(k+1)!-1+(k+1)(k+1)!$
$=(k+1)!(1+k+1)-1$
$=(k+1)!(k+2)-1$
$=(k+2)!-1$

which is what you needed to show, so the statement is true.