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?
 A: $$ 
\begin{align} 
\color{red}{0}\lt \color{blue}{x}\le\color{green}{1} \quad &\rightarrow\quad 2^\color{red}{0}\lt2^\color{blue}{x}\le2^\color{green}{1} \quad\Rightarrow\quad x\le2^0\lt2^x \\[2mm] 
\color{red}{1}\lt \color{blue}{x}\le\color{green}{2} \quad &\rightarrow\quad 2^\color{red}{1}\lt2^\color{blue}{x}\le2^\color{green}{2} \quad\Rightarrow\quad x\le2^1\lt2^x \\[2mm] 
\color{red}{2}\lt \color{blue}{x}\le\color{green}{4} \quad &\rightarrow\quad 2^\color{red}{2}\lt2^\color{blue}{x}\le2^\color{green}{4} \quad\Rightarrow\quad x\le2^2\lt2^x \\[2mm] 
\color{red}{4}\lt \color{blue}{x}\le\color{green}{16} \,\,\,\, &\rightarrow\quad 2^\color{red}{4}\lt2^\color{blue}{x}\le2^\color{green}{16} \,\,\,\Rightarrow\quad x\le2^4\lt2^x \\[4mm] 
\cdots\quad\text{etc}\quad &\rightarrow\quad \color{red}{x\lt2^x\quad\colon x\in\mathbb{R}}
\end{align} 
$$
A: Note: Hazem Orabi's solution is really the one I prefer personally but since i had fun messing with a more convoluted and ultimately inconclusive approach I'd just like to share it. 
Solution
Let's do this for rational numbers $x = p/q$ and then just assume it extends properly to real numbers.
Let $p,q$ be relatively prime and let's just note that
$$0 < x < 2 \Rightarrow p < 2q$$
Now the number $2^{x} = 2^{p/q}$ is the solution to the equation $y^q = 2^p$ and the idea that $2^x > x$ can be rephrased as the implication 
$$p < 2q \quad \text{and} \quad y^q = 2^p \quad \Rightarrow \quad qy > p \quad \text{(to be proven)}$$
The reason I phase it like this is that we're just dealing with integral powers the results of which are hopefully more intuitive (easier to prove) than rational or irrational powers. 
Let us start by observing 
$$(qy)^q = q^q y^q = q^q 2^p$$
And then divide our intention to two special cases $q < p$ and $p < q$
Case 1. $p < q$. In this  case we use $q^q > p^q$
$$(qy)^q = q^q 2^p > p^q 2^p > p^q  $$
and therefore
$$(qy)^q > p^q \Leftrightarrow qy > p$$
Case 2. $q < p$. In this  case we use $2^p > 2^q$
$$(qy)^q = q^q 2^p > q^q 2^q > (2q)^q > p^q   $$
(where $2q > p$ was invoked in the final step) and therefore again 
$$(qy)^q > p^q \Leftrightarrow qy > p$$
(The principle invoked over and over again is that integral powers preserve order which while not entirely trivial is at least natural)
A: 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 $2^x>x$.
