Prove $\sum_{i=1}^n i! \cdot i = (n+1)! - 1$? Prove the summation: 
$$\sum_{i=1}^n i! \cdot i = (n+1)! - 1$$
using induction. 
base case: 
$n=1$:
\begin{align*}
\sum_{i=1}^1 i! \cdot i &= (1+1)! - 1 \\
1 &= 2 - 1 \\
1 &= 1
\end{align*} 
This is a question from my test review packet, currently have the base case completed and I am a bit lost on where to go from there. Any help/hints are appreciated.
 A: Perhaps some of the "tricky" algebra is what is really tripping you up: you have shown that the base case holds. Now you assume that the statement is true for some fixed $k\geq 1$ (i.e., $\color{blue}{\sum_{i=1}^k i!\cdot i=(k+1)!-1}$), and your goal is to use this assumption (called the inductive hypothesis) to show that
$$
\color{green}{\sum_{i=1}^{k+1}i!\cdot i=(k+2)!-1}
$$
holds. Start with the left-hand side and work your way to the right-hand side:
\begin{align}
\color{green}{\sum_{i=1}^{k+1}i!\cdot i} &= \color{blue}{\sum_{i=1}^k i!\cdot i}+(k+1)!\cdot(k+1)\tag{by defn. of $\Sigma$}\\[0.5em]
&= \color{blue}{(k+1)!-1}+(k+1)!\cdot(k+1)\tag{by inductive hypothesis}\\[0.5em]
&= (k+1)!(1+k+1)-1\tag{factor out $(k+1)!$}\\[0.5em]
&= (k+1)!(k+2)-1\tag{simplify}\\[0.5em]
&= \color{green}{(k+2)!-1}.\tag{by defn. of factorial}
\end{align}
Can you see how this proves your claim?
A: Use induction!  By the inductive hypothesis,
$$
\sum_{i=1}^{n+1} i! i = \sum_{i=1}^{n} i! i + (n+1)!(n+1) = [(n+1)! - 1] + (n+1)!(n+1)
$$
Now just finish off the righthand side.
