Prove by Mathematical Induction: $1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$

Your LHS may not look much like your RHS yet, but that's because you haven't finished getting it into the simplest possible form. You have $(k+1)k! - 1 + (k+1)^2 k!$. You're looking to get something minus $1$, so that's somewhat promising. Now what factors do the other two terms (the ones involving $k$) have in common?

It's a special case of telescopic induction. This post has a very short and simple inductive proof of

Theorem $\rm\displaystyle\,\ \sum_{i\,=\,1}^n f(i) = g(n)\iff g(1) = f(1)\ {\rm and}\ g(n\!+\!1)-g(n) = f(n\!+\!1)\ $ for $\,n \ge 1.$

Applied to your case, where $\rm\,f(n) = n n!\:$ and $\rm\:g(n) = (n+1)!-1,\,$ we have $\rm\: g(1)=1 = f(1),\:$ and

$$\begin{eqnarray}\rm g(n\!+\!1)-g(n) &=&\rm (n\!+\!2)!-1-((n\!+\!1)!-1) \\ &=&\rm (n\!+\!2)!\,-\,(n\!+\!1)! \\ &=&\rm (n\!+\!2 -1)(n\!+\!1)! \\ &=&\rm (n\!+\!1)(n\!+\!1)! \\ &=&\rm f(n\!+\!1) \end{eqnarray}$$

That completes the proof using the theorem. This method works quite widely for inductive proofs involving sums and products. You can find many more examples of telescopy and related results in other answers here.

$$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!) = (n+1)!-1$$ then $$1(1!) + 2(2!) + \cdot \cdot \cdot +n(n!)+(n+1)(n+1)! =(n+1)!-1+(n+1)(n+1)!=$$ $$(n+1)!(n+1+1)-1=(n+2)!-1=((n+1)+1)!-1$$