Use of Second Principle of Induction
\[1 \times 1! +2 \times 2!+...+n \times n!\]
for all \[n \ge 0\]
using induction.Let
\[P(k)\]
be the statement that the above statement is true. Then\[0 \times 0!=(0+1)!-1\]
Both sides equal zero so
\[P(0)\]
is true.Suppose that
\[P(1), \; P(2),..., \; P(k)\]
are true (Second Principle of Mathematical Induction)\[\begin{equation} \begin{aligned} (1 \times 1! &+ 2 \times 2!+...\\ &+ k \times k!)+(k+1) \times (k+1)! \\ &= (2!-1)+((3!-1)-(2!-1))+... \\ &+ (((k+2)!-1)-((k+1)!-1)) \\ &=(k+2)!-1 \end{aligned} \end{equation}\]
Hence
\[P(k+1)\]
is true and the identity is proved. 
