Prove the inequality: $n^{n+1}>(n+1)^n,(n\ge 3)$ I know how to prove it by induction.
Is it possible to use mean value theorem?
$$f(n)=n^{n+1}-(n+1)^n$$
$$f(0)=0$$
$$f^{'}(n)>f(0)\Rightarrow f(n)>f(0)$$
This is correct for starting value $n=0$.
Does it mean that from the condition $(n\ge 3)$ inequality doesn't hold using M.V.T? 
 A: It's possible. To summarize the proof you have that $f(a)=0$ and $f'(x)>0$ for $x>a$, then by MVT if $f(b)\le 0$ for $b>a$ then you would have a $\xi\in(a,b)$ such that $f'(\xi) = {f(b)-0\over b-a} < 0$ which is a contradiction. Therefore $f(x) > 0$ for all $x>a$.
The rest is mostly a matter of showing that $f'(x)>0$ for $x>a$:
Let's instead consider $(n+1)\ln(n) = \ln(n^{n+1}) > \ln((n+1)^n) = n\ln(n+1)$. Now let $f(x) = (x+1)\ln(x) - x\ln(x+1)$. Now the derivative is
$f'(x) = \ln(x) + {x+1\over x} - \ln(x+1) - {x\over x+1} = \ln{x\over x+1} + {x+1\over x} - {x\over x+1}$
now put $t = {x\over 1+x}$ and we get $f'(x) = \ln t + 1/t - t$, now if $x > 0$ we have $0 < t < 1$ so $f'(x) > 0$.
To see these later claims you can do the same kind of reasoning over one time more, let $g(t) = \ln t + 1/t - t$, $g'(x) = {1\over t} - {1\over t^2} -1 = -({1 \over t}-1/2)^2-{3\over4}$, and $g(1) = 0$ so if $f(t) \ge 0$ for some we would find $g'(\xi)$ to be negative which it isn't. Similar reasoning can go for ${x \over 1+x}$. 
