Prove that $n^n>\left(\dfrac{n+1}{2}\right)^{n+1}$ for all positive integer $n>1$. Question: Prove that $n^n&gt;\left(\dfrac{n+1}{2}\right)^{n+1}$ for all positive integer $n&gt;1$.   

I could not understand what should be the initial approach.
We want to show the $2$s at the bottom are enough to defeat the $+1$s at the top. The effect of the $2$s is easy. Just write 

$$ \left(\dfrac{n+1}{2}\right)^{n+1} = \frac{1}{2^{n+1}}(n+1)^{n+1}$$
We'd like a bound like  $(n+1)^{n+1} &lt; Cn^n$ for some $C &lt; 2^{n+1}$. 

$$(n+1)^{n+1}= (n+1)(n+1)^{n}$$

To bound $(n+1)^n$ use the binomial expansion, round each $n^i$ up to $n^n$, then use how the coefficients add up to $2^n$. This should complete the proof.
We want to show the $2$s at the bottom are enough to defeat the $+1$s at the top. The effect of the $2$s is easy. Just write 

$$ \left(\dfrac{n+1}{2}\right)^{n+1} = \frac{1}{2^{n+1}}(n+1)^{n+1}$$
We'd like a bound like  $(n+1)^{n+1} &lt; Cn^n$ for some $C &lt; 2^{n+1}$. 

$$(n+1)^{n+1}= (n+1)(n+1)^{n}$$

To bound $(n+1)^n$ use the binomial expansion, round each $n^i$ up to $n^n$, then use how the coefficients add up to $2^n$. This should complete the proof.