Proof that $\forall x\in (0,\frac{1}{e}),(x^x >e^\frac{-1}{e})$ Let $f:(0,\infty)\to\mathbb{R}:x\mapsto x^x$
Prove, using the mean value theorem, that:
$$\forall x\in (0,\frac{1}{e}),(f(x) >e^\frac{-1}{e})$$
Mean Value Theorem
Let $f$ be a continuous function on $[a,b]$ that is differentiable on $(a,b)$. Then there exist [at least one] $x$ in $(a,b)$ such that 
$$f'(x)=\frac{f(b)-f(a)}{b-a}$$
I think I can prove that $f$ is continuous on $[0,\frac{1}{e}]$ and differentiable on $(0,\frac{1}{e})$. By the mean value theorem:
$$\exists x\in (0,\frac{1}{e}),f'(x)=\frac{f(\frac{1}{e})-f(0)}{\frac{1}{e}-0}=\frac{(\frac{1}{e})^\frac{1}{e}}{\frac{1}{e}}=e\cdot(e^{-\frac{1}{e}})$$
 A: You can't use MVT in $[0,\frac{1}{e}]$ as $f$ is not defined at $0$ (we need continuity there for MVT). You can avoid this shortcoming by extending $f$ continuously. Because
$$\lim_{x\to 0^+}x^x=\lim_{x\to 0^+}e^{x\ln x}=1$$
the function:
$$g(x)=\begin{cases}
x^x &\mbox{if } x\in (0,\frac{1}{e}]\\
1&\mbox{if } x=0\end{cases}$$
is continuous and differentiable in $(0,\frac{1}{e})$ with derivative 
$$g^{\prime}(x)=x^x(\ln x+1)$$
If you apply MVT on $[0,\frac{1}{e}]$ you won't get any good results as you have seen. Instead you will get
$$\xi^\xi =e^{\frac{-1}{e}+1}-1$$ 
for some $\xi\in (0,\frac{1}{e})$ which is nothing like what you want to show. Instead you should apply MVT in this interval: $[x,\frac{1}{e}]$.
Here is what I propose:
Since $x\in (0,\frac{1}{e})$, $\ln x<-1$ and
$$f^{\prime}(x)<0$$
If we apply the MVT now on $[x,\frac{1}{e}]$ (for fixed $0<x<\frac{1}{e}$) we have that $\exists \xi \in (x,\frac{1}{e})$ so that
$$\frac{e^{\frac{-1}{e}}-x^x}{\frac{1}{e}-x}=f^{\prime}(\xi)<0\Rightarrow e^{\frac{-1}{e}}<x^x$$
Note that in this approach you don't need to define $g$ as $f$ is continuous in $[x,\frac{1}{e}]$ for fixed $0<x<\frac{1}{e}$
Moral: When you want to prove an inequality using MVT the interval that you should use may be of the form $[a,x]$ or $[x,b]$
A: Take $log$, it is equivalent to prove $xln(x)>-\frac{1}{e}$. Using the points $x, \frac{1}{e}$, then mean value theorem tells us that $\frac{xlnx+\frac{1}{e}}{x-\frac{1}{e}}=1+ln(a)$, for some $a\in [x, \frac{1}{e}]$, so we need to show $(\frac{1}{e}-x)(1+ln(a))<0$, which is clear since $\frac{1}{e}>x>0, ln(a)<ln(\frac{1}{e})<-1$
