What Values of $a$ Satisfy $a^x>x$ For All Real $x?$ This question can be asked in two different (but equivalent) ways:
$1.$ What values of $a$ satisfy $a^x>x\forall x\in\mathbb{R}?$
$2.$ Find the interval in which $a$ has to lie so that the graph of $f(x)=a^x$ is always above the graph of $\textrm{id}(x)=x.$
My Attempt:
Clearly, $a$ cannot be negative. Any negative number raised to an odd positive integer is negative, and hence is less than the exponent. More explicitly, $(-1)^1=-1<1.$
$a=0$ does not work either. A counterexample to demonstrate this is $0^3=0<3.$
After working out that $a>0$ using the above arguments, I thought that all $a>1$ would satisfy $a^x>x,$ but that's clearly untrue. For example, the graph $f(x)=1.3^x$ cuts the graph of $\textrm{id}(x)=x$ at $x_1≈1.471,$ and $x_2≈7.857.$ Moreover, between these two values, the graph of $f$ lies below the graph of $\textrm{id}.$ I have that $1.444667<a≤1.444668.$ I got this using desmos. According to desmos, the graph of $f(x)=1.444667^x$ intersects the graph of $\textrm{id}(x)=x$ at two points while $f(x)=1.444668^x$ does not intersect $\textrm{id}(x)=x$ at all. Is there a closed form for $a?$ I have attached screenshots of the graphs below.

Fig. $1.$ Graphs of $f(x)=1.444668^x,$ and $\textrm{id}(x)=x.$

Fig. $2.$ Graphs of $f(x)=1.444667^x,$ and $\textrm{id}(x)=x.$

Fig. $3.$ Fig. $2,$ but zoomed in.
Note: While the screenshot may make it seem like the two graphs in Fig. $1$ intersect, they don't (atleast according to desmos).
 A: The $a$ you seek is $e^{1/e}$.
Indeed, $a^x>x$ for all $x$ iff $a > x^{1/x}$ for all $x$.
Therefore, the minimum $a$ is the global maximum of $x^{1/x}$.
The function $f(x)=x^{1/x}$ has a global maximum at $x=e$ and so $a^*=e^{1/e}$
A: The limiting case is
$$\large a=e^{\frac 1e}=1.444667861009766133658339$$
Now, prove it (this is not difficult).
Edit
Since, at the time I write this edit, ou know the steps, make the problem more general, that is to say
$$a^{x^b} > x$$
So, repeating the steps given in the other answers, you need to solve for $a$ and $x$ the two equations
$$a^{x^b} = x \tag 1$$
$$b \log (a) x^{b-1} a^{x^b}=1\tag 2$$
The ratio of $(2)$ to $(1)$ gives
$$b \log (a) x^{b}=1 \quad \implies \quad x=(b \log (a))^{-1/b}$$ and, back to $(1)$
$$a=e^{\frac{1}{e b}}\quad \implies \quad x=e^{\frac{1}{ b}}$$
A: The question is equivalent to finding $a\in \mathbb{R}$ for which $a^x-x>0$ for all $x\in \mathbb{R}$. Let's try finding the minimal possible value for $f(x)=a^x-x$. Differentiating and equating to $0$:
$$f'(x)=a^x\cdot \ln(a)-1=0 \implies a^x\cdot \ln(a)=1\implies a^x=(\ln(a))^{-1}$$
$$\implies x=\log_a ((\ln(a))^{-1})$$
This is a continuous function, and it is clear that it has a minimum. We can confirm this by checking the second derivative at $x$ for example, but I think understanding that the vertical distance between the graphs has a minimum is enough.
So, this $x$ gives us the minimal value for $f$:
$$f(\log_a ((\ln(a))^{-1}))=a^{\log_a ((\ln(a))^{-1})}-\log_a ((\ln(a))^{-1})=\frac{1}{\ln(a)}+\frac{\ln(\ln(a))}{\ln(a)}$$
We would like this minimal value to be $0$, so we get
$$1+\ln(\ln(a))=0\implies \ln(a)=e^{-1}\implies a=e^{\frac{1}{e}} (\approx1.44466786...)$$
