minimum number $\alpha$ such that for all $x$: $\alpha^x \geqslant x$ This is a short question which I'm asking just out of curiosity:

Find the minimum positive number $\alpha \in \mathbb{R}$ such that $\forall x \in \mathbb{R}: \alpha^x \geqslant x$.

Maybe somebody even knows how it relates to other mathematical constants, e.g. to $e$.
You're also welcome with some hints or references to the corresponding literature :).

This is only to summarize the answers:
The inequality holds for all $x \leqslant 0$ irrespectively to $\alpha$.
Assuming $x > 0$ we can apply the following transformations:
\begin{gather*}
\alpha^x > x \\
x\ln{\alpha} > \ln{x} \\
\ln{\alpha} > \frac{\ln{x}}{x} \\
\alpha > e^{\frac{\ln{x}}{x}}
\end{gather*}
If we want $\alpha$ to be greater than $e^{\frac{\ln{x}}{x}}$ for any $x$ we need to make it greater than maximum of $e^{\frac{\ln{x}}{x}}$.
This maximum is some number $M$ and we need to find minimum possible $\alpha$ such that $\alpha > M $.
This is impossible for real numbers $\mathbb{R}$ because for any number $\alpha$ we can find infinitely many numbers $\beta$ such that $M < \beta < \alpha$ - there is no minimum possible number greater than $M$. That is why J.G. was writing that my question was posed incorrectly and that I needed to replace $a^x > x$ with $a^x \geqslant x$.
Let's fix this
\begin{gather*}
\alpha^x \geqslant x \\
\ldots \\
\alpha \geqslant e^{\frac{\ln{x}}{x}}
\end{gather*}
Now we can try to find $M$:
\begin{gather*}
\left(e^{\frac{\ln{x}}{x}}\right)' = 0 \\
\left(\frac{\ln{x}}{x}\right)' = 0 \\
1 - \ln{x} = 0 \\
x = e;
\end{gather*}
which gives
\begin{equation*}
M = e^{\frac{\ln{e}}{e}} = e^{\frac{1}{e}}
\end{equation*}
The number $\alpha$ which we are seeking is the minimum possible number $\alpha \geqslant M$ - this is indeed $M$.
Therefore
\begin{equation*}
\alpha = e^{\frac{1}{e}}
\end{equation*}
