13.2. Calculate the sum

$$
1 \cdot 1!+2 \cdot 2!+3 \cdot 3!+\ldots+n \cdot n!
$$
13.2. We will prove by induction on $n$ that the considered sum equals $(n+1)!-1$. For $n=1$, we get the obvious equality $1 \cdot 1! = 2! - 1$. Suppose the required equality is proven for some $n$. Then

$$
\begin{aligned}
1 \cdot 1! + 2 \cdot 2! + \ldots + n \cdot n! & + (n+1) \cdot (n+1)! = \\
& = (n+1)! - 1 + (n+1) \cdot (n+1)! = (n+2)! - 1
\end{aligned}
$$

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