When $n=1$, the left side equals $1$ and the right side equals $2!-1=1$. Hence, the relation holds. Assume that the relation holds when $n=k$ where $k \ge 1$, i.e. $$1\cdot 1! + 2\cdot 2! + \cdots + k\cdot k!=(k+1)!-1$$ Then when $n=k+1$: \begin{align} & 1\cdot 1! + 2\cdot 2! + \cdots + k\cdot k! + (k+1)\cdot (k+1)!\\ & = ((k+1)! - 1) + (k+1)\cdot(k+1)!\\ & = (k+1)! - 1 + (k+1)\cdot(k+1)!\\ & = (k+1)!(1+(k+1)) - 1\\ & = (k+2)!-1 \end{align} Therefore, by the principle of mathematical induction, the relation holds for all $n \ge 1$.