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I need to evaluate the following

$$\frac{(n+1)!}{(n+1)^{(n+1)}} * \frac{n^n}{n!}$$

It should come to $$(\frac{n}{n+1})^n$$

Currently, I only know that the $(n+1)!$ cancels with the $n!$ to make $n+1$.

But, how would I evaluate the remaining?

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$$\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}$$ $$ \frac{n+1}{(n+1)^{(n+1)}}n^n $$ $$ \frac{n+1}{(n+1)^{n}(n+1)}n^n $$ $$ \frac{1}{(n+1)^{n}}n^n $$ $$ \left(\frac{n}{n+1}\right)^n $$

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  • $\begingroup$ Excellent, got it! Thanks! $\endgroup$ – Mr Croutini Dec 21 '13 at 17:06
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VERY BIG HINT

Separate the exponent by using the exponent rule $a^{m+n} = a^ma^n$.

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$$\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}=\frac{(n+1)n!}{(n+1)(n+1)^{n}}\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n$$

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(Big) Hint: $$\frac{a^x}{a^y}=\frac1{a^{y-x}}$$ for all real $a,x,y$ with $a>0.$

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