**e**^{π} > π^{e}.
The underlying idea here is that a^{b} > b^{a} whenever b > a, given that a and b are both greater than 1.
If you look at the graphs of y=x^{a} and y=a^{x}, they will cross, of course, at x=a.
For higher values of x (again, given that a>1), the graph of a^{x} rises faster than x^{a}.
Thus, when x>a, a^{x} > x^{a}.
1 < e < π, so e^{π} > π^{e}.
Another way to prove this directly follows:
π > e, so ln(π) > 1
e^{(1-1)} = 1
e^{(x-1)} > x (for x>1)
e^{[ln(π)-1]} > ln(π)
(e^{ln(π)})/e > ln(π)
e^{ln(π)} > e ln(π)
π > e ln(π)
π > ln(π^{e})
e^{π} > e^{ln(πe)}
e^{π} > π^{e} |