How can I

compare (without calculator or similar device) the values of \pi^e and e^\pi ?

**Answer**

Another proof uses the fact that \displaystyle \pi \ne e and that e^x > 1 + x for x \ne 0.

We have e^{\pi/e -1} > \pi/e,

and so

e^{\pi/e} > \pi.

Thus,

e^{\pi} > \pi^e.

**Note**: This proof is not specific to \pi.

**Attribution***Source : Link , Question Author : Mirzodaler , Answer Author : David*