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I'm unsure on how this simplifies. Could anyone explain how and what technique is used to produce the answer? Any help would be most appreciated. $$ (n+1)(n+1)!+(n+1)!−1 \\ = (n + 1)!((n + 1) + 1) − 1 \\ = (n + 1)!(n + 2) − 1 $$

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    This is just the distributive law...$AB+B-1=B(A+1)-1$. – 2017-01-17
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    @lulu Why don't you just expand what you have there with a sentence or two and post it as an answer? – 2017-01-17
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    Presumably, you were disturbed by the factor reversal. $(n+1)(n+1)!+(n+1)!=((n+1)+1)(n+1)!=(n+2)(n+1)!$. – 2017-01-17
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    Final answer is $(n+2)! - 1$. =) – 2017-01-17

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This is simply an application of the distributive law for addition and multiplication. It looks worse because the terms are themselves a little complicated. To clarify matters, let $A=(n+1),B=(n+1)!$. Then your expression is simply $$AB+B-1$$

And this can be rewritten as $$B(A+1)-1$$

That's all that's going on here!

Worth pointing out: One could take it one step further and remark that $(n+2)(n+1)!=(n+2)!$ thus your expression becomes $$(n+2)!-1$$

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    thank you very much for your responce. I understand it now :) – 2017-01-17