Question

I need to prove 1*1! + 2*2!+...+n*n!=(n+1)!-1 with mathematical induction

Answer 1

for n=1,
RHS=(1+1)!-1=2-1=1=LHS.
therefore true for n=1,

assume true for n=p,
$$1\times1!+2\times2!+...+p\times p!=(p+1)!-1$$
add (p+1)\times (p+1)! to both sides,
$$\begin{align*}
1\times1!+2\times2!+...+p\times p!+(p+1)\times(p+1)!&=(p+1)!-1+(p+1)\times (p+1)!\\
&=(p+1)!\left[1+p+1\right]-1\\
&=(p+1)!\times (p+2)-1\\
&=(p+2)!-1\\
&=[(p+1)+1]-1
\end{align*}$$
therefore it is true for n=p+1