I know two solutions, both use ideas from calculus.

Solution 1: Set up the function f(x)=ex/xe, over the domain x>0; we wish to know how f(π) compares with 1.

Using the quotient rule, calculate f ′ (x)=(ex/xe)(1-e/x), which is negative for 0 < x < e, zero at x=e, and positive for x > e.

By the first derivative test, f reaches a unique minimum at x=e, in which case f(e)=1. Thus f(x) > 1 for all other values of x; in particular, for x=π. Thus eπ > πe.

Solution 2: Since the exponential function equals its own derivative, the equation for the straight line tangent to the curve y=ex at (0,1) is y = 1+x
(the first two terms in the Maclaurin series). Since the curve is concave up, the line lies strictly below the curve, except at x = 0.
Thus we have the strict inequality ex > 1+x for all x ≠ 0. Now fix x=(π /e)-1. Since π ≠ e, we know x ≠ 0.

By plugging this value of x into the inequality, multiplying both sides by e, and then raising both sides to the power of e
(the function y=xe, x > 0, is a strictly increasing function), we obtain eπ > πe.

Comment: Both solutions actually give you more than just the stated inequality; namely they give you ea > ae for any positive a ≠ e.