Exercise 1044Logic, sets and proofs › Proof techniques › InductionL1Added 3 Nov 2026

A factorial identity

#induction#factorial#identity

Statement

Prove that $\sum_{k=1}^{n}k\cdot k!=(n+1)!-1$ .

Hint

$k\cdot k!=(k+1)!-k!$ , telescoping.

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Identity. $k\cdot k!=(k+1-1)k!=(k+1)!-k!$ .

Sum. $\sum_{k=1}^n k\cdot k!=\sum_{k=1}^n[(k+1)!-k!]=(n+1)!-1!$ .

Induction alternative. Base case $n=1$ : $1=2!-1=1$ . ✓ Inductive step: $\sum_{k=1}^{n+1}k\cdot k!=(n+1)!-1+(n+1)(n+1)!=(n+1)!(n+2)-1=(n+2)!-1$ .

Conclusion. The identity holds.