Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 2 down vote accepted

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

share|cite|improve this answer
    
Excellent, got it! Thanks! – Mr Croutini Dec 21 '13 at 17:06

VERY BIG HINT

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

share|cite|improve this answer

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

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.