Prove that 1·1 ! + 2·2!+· ··+ n · n! = (n+I)!-1 whenever n is a positive integer.

$\begin{aligned} &\text { To proof: } 1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !=(n+1) !-1 \text { for every positive integer } n . \\ &\text { PROOF BY INDUCTION: } \\ &\text { Let } P(n) \text { be } 1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !=(n+1) !-1 \\ &\text { Basis step } n=1 \\ &1 \cdot 1 !+2 \cdot 2 !+\ldots+n \cdot n !=1 \cdot 1 !=1 \cdot 1=1 \\ &(n+1) !-1=(1+1) !-1=2 !-1=2-1=1 \\ &\text { We then note } P(1) \text { is true. } \\ &\text { Induction step: Let } P(k) \text { be true. } \\ &1 \cdot 1 !+2 \cdot 2 !+\ldots+k \cdot k !=(k+1) !-1 \\ &\text { We need to prove that } P(k+1) \text { is also true. } \end{aligned}$

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

We then note that $P(k+1)$ is also true.

Conclusion By the principle of mathematical induction, $P(n)$ is true for all positive integers $n$ .

Thus, $1·1 ! + 2·2!+· ··+ n · n! = (n+1)!-1$ whenever n is a positive integer.

Hence Proved