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This question already has an answer here:

How do I go about simplifying this:

$$\sum_{k=1}^nk \cdot k!$$

Wolfram alpha tells me it's the same as $(n+1)!-1$ but I don't see how.

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marked as duplicate by Brian Borchers, Nosrati, KonKan, José Carlos Santos, Stefan4024 Dec 1 '17 at 23:19

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ If true it must follow by induction. $\endgroup$ – David C. Ullrich Dec 1 '17 at 16:18
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Hint: $$k\cdot k! = [(k+1)-1]\cdot k! = (k+1)!-k!$$

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