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 $a\geq1$.
First, observe that \begin{align} &(1 + x)^{(1 + x)} > e^x \\&\iff (1 + x) \log (1 + x) > x \\&\iff (1 + x)\log(1 + x) - x > 0 \end{align} Set $f(x) = (1 + x)\log(1 + x) - x$. Then $$ f'(x) = \log(1 + x) + 1 - 1 = \log(1 + x) $$ So $f'(x) > 0$ whenever $x > 0$. That is, $f$ is strictly increasing on $(0, \infty)$. On the other hand $$ f(0) = (1 + 0) \log(1 + 0) - 0 = 0 $$ These two facts together show that $f(x) > 0$ whenever $x > 0$, which is equivalent to what needed to be shown.
Hint: apply $\ln$ to both sides of the inequality.
$$(1+x)\ln(1+x) > x\Rightarrow (1+x)\ln(1+x) - x > 0$$
The proof for $x \geq e-1$ is straightforward. For $0 < x < e-1$, by the concept of derivative show that it is increasing in the specified area.
Most such inequalities seem to follow from the familiar result that $e^t > 1 + t$ for all real $t \ne 0$.
In this case, putting $t = -\log(1 + x)$: $$ (1 + x)\log(1 + x) = e^{-t}(-t) > e^{-t}(1 - e^t) = e^{-t} - 1 = x \quad (x > -1,\ x \ne 0). $$
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 $x>0 \iff a\ge 1$.
$$ (1+x)^{1+x} = 1 + x + x^2 + \frac{x^3}{2} + \frac{x^4}{3} + \cdots $$ is clearly greater than $$ e^{x} = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{24} + \cdots $$ for $x > 0.$
