Show that $(1+x)^{(1+x)}>e^x$ How can I prove that $(1+x)^{(1+x)}>e^x$ for all $x>0$?
The problem arose as I tried to prove the well-known & intuitive econometric principle that the more often you compound your interest, the more interest you ultimately get (in maths, that $\frac{d}{dn}((1+\frac{1}{n})^n)>0$ for $n>0$).
An interesting further problem is to prove that $(a+x)^{(a+x)}>e^x$ is true for all $x>0$ if and only if $a\geq1$.

Your general case is easy with this corollary of the Mean value theorem:

Let $f,g$ be two differentiable functions defined on an interval $I$ and $x_0\in I$.
If $f(x_0)\ge g(x_0)$ and $f'(x)>g'(x)$ for all $x>x_0,\:x\in I$, then  $f(x)>g(x)$ for all $x>x_0,\:x\in I$.

Indeed it is enough to compare the logs: set $f_a(x)=(a+x)\ln(a+x)$, $g(x)=x$. We have $f(0)=a\ln a\ge g(0)=0$ iff $a\ge 1$.
Now compare the derivatives:
$f'_a(x)=\ln(a+x)+1>1$ for all $x>0\iff a+x>1$ for all $x>0 \iff a\ge 1$.
