Prove that $x<2^x$ 
How can we prove that given that $x\leq 2$, then $2^x>x$?

I think that this seems to be intuitively correct but I don't know how to prove it. Can it be proven without calculus?

The inequality is obvious for $x\le0$, so we can assume $x>0$.
Consider the function $f(x)=x\log 2-\log x$ (natural logarithm). It's easy to prove that this function has a minimum for $x=1/\log2$ and that
$$
f(1/\log2)=1+\log\log2
$$
Since $2^2>e$, we have $2\log2>1$, so $\log2>1/2$ and
$$
\log\log2>-\log2
$$
Therefore $1+\log\log2>1-\log2>0$, because $2<e$.
Since the minimum of $f$ is positive, we have $x\log2>\log x$, for every $x>0$, hence $2^x>x$.
