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Possible Duplicate:
Summation of a factorial

This equation is given:
$$ 1\cdot1! + 2\cdot2! + 3\cdot3! + \ldots + n\cdot n! = (n+1)! - 1 $$

I've solved it using mathematical induction but I'm curious what could be the other possible ways to prove it.

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    @MattN. It's *not* a dup since the OP asks for ways *other* than induction, whereas the linked question asks for proofs by induction.2012-03-15
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    @Matt: It’s definitely not a duplicate, since that one wanted an inductive argument, and this one explicitly doesn’t.2012-03-15
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    Sorry for being somewhat too quick on the trigger.2012-03-15
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    @BrianM.Scott the [first answer](http://math.stackexchange.com/a/18580/2468) in the question Matt linked contains an non-inductive proof.2012-03-15
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    There cannot be a non-inductive proof, since the sum is defined recursively. As a rough guide, if it has "$\dots$" it requires induction, unless it is a consequence of a prior result that required induction. But many inductive proofs are so transparent that we do not make the induction explicit.2012-03-15
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    @tendua The accepted answer *does* use induction (hidden in the ellipses). It is nothing but the standard inductive proof expressed less rigorously.2012-03-15
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    @tendua My point has nothing at all to do with issues about acceptance. Rather, I am trying to get you to think more deeply about mathematical induction.2012-03-15
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    It bears questioning what definition of "prove" is being used here. It's perfectly appropriate to state that this equation can be proven by analogy, though its dependence on unbounded integer $n$ should require an inductive proof one or more orders below this given analogy.2012-03-15
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    @All: Does closing the question means that there couldn't be proofs without induction and hence there should be no new answers so the question isn't open for discussion anymore?2012-03-16
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    @tendua No one got pinged by your "@All" ping, you need to ping using names. Sorry for the late reply. I voted to close your question because I thought the question I linked to answered your question. Now if you think your question is not answered and you'd like it to be reopened let us know. We can always vote to reopen it. Hope this helps. If you'd like to discuss any of the answers or anything else feel free to drop by the main chat room. Cheers2012-03-16
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    @MattN.: thanks for the valuable information.2012-03-16
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    @tendua [Here](http://math.stackexchange.com/questions/120674/must-we-use-induction-to-prove-a-statement-for-all-integers)'s a follow up question to BD's answer. Someone just pointed it out to me in chat.2012-03-16

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