TITLE: Why $e^x$ is always greater than $x^e$?
QUESTION [18 upvotes]: I find it very strange that $$ e^x \geq x^e \, \quad \forall x \in \mathbb{R}^+.$$ 
I have scratched my head for a long time, but could not find any logical reason. Can anybody explain what is the reason behind the above inequality? I know this is math community, but I'd appreciate a more intuitive explanation than a technical one.

REPLY [3 votes]: I'd appreciate a more intuitive explanation than a technical one. 

Well on that, try to consider the behaviour of $2^x$.  Every time I increase x by one I get another (binary) digit out of my expression.  So I quickly get a huge number.  If I double $x$ I get way more digits.
Now compare that with $x^2$.  Every time I increment $x$ I will not generally get an extra digit from this expression.  In fact as $x$ gets really large there may be no difference in the number of digits between two "adjacent" values of the expression.
For example the number of binary digits required to represent $1000^2$ is the same as the number to represent $1001^2$.  In fact it's 10 binary digits for both.  But $2^{1000}$ requires one thousand binary digits.  So a huge difference in growth.
So we see an exponentiation function grows much faster than the square function.
If binary digits aren't your "thing" then use base $10$ and compare $10^x$ and $x^{10}$.  It's the same argument.
Now if you consider your specific request about $e^x$ and $x^e$ you should see, intuitively, that this is a similar situation in terms of relative growth.  We no longer have the convenience of binary digits, but it's pretty clear the same rationale holds.