The equation $a^x=x$ for $a>1$. The following problem arises when I was sketching the inverse of $f(x)=a^x$ graphically for $a>1$:
1) Is there an $a>1$ such that the equation $a^x=x$ has a unique solution $x\in\mathbb{R}$?
2) If so, then how do we find such $a$ explicitly if possible? 
The answer to the first question seems to be yes, I tried solving the equation $a^x=x$ for $a=1.4$ and $a=1.5$, which has two and no solutions respectively; but I personally would love to see a proof without geometry argument. 

I guess I will add my two cents 
$$a^x=x$$
$$a=x^{{1}/{x}}$$
now 
$$e^{x}\ge 1+x$$
$$e^{(x-e)/{e}}\ge x/e$$
$$e^{x/e}\ge x$$
$$e^{1/e} \ge x^{1/x}$$
Therefore maximum of $a=e^{1/e}$ for solution. Thus there exists solution for $a\in\mathbb{(1,e^{1/e})}$
