# How do these equate?

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?

$$\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}$$ $$\frac{n+1}{(n+1)^{(n+1)}}n^n$$ $$\frac{n+1}{(n+1)^{n}(n+1)}n^n$$ $$\frac{1}{(n+1)^{n}}n^n$$ $$\left(\frac{n}{n+1}\right)^n$$
Separate the exponent by using the exponent rule $a^{m+n} = a^ma^n$.
$$\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}=\frac{(n+1)n!}{(n+1)(n+1)^{n}}\frac{n^n}{n!}=\frac{n^n}{(n+1)^n}=\left(\frac{n}{n+1}\right)^n$$
(Big) Hint: $$\frac{a^x}{a^y}=\frac1{a^{y-x}}$$ for all real $a,x,y$ with $a>0.$