Recall that: e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots
Therefore, \forall x > 0 , e^x > 1 + x .
Substituting x = \frac{\pi}{e} - 1 in the above inequality, we get:
e^{\frac{\pi}{e} - 1} > \frac{\pi}{e} \implies e^{\frac{\pi}{e}} > \pi \implies e^{\pi} > \pi^{e}
Therefore, \forall x > 0 , e^x > 1 + x .
Substituting x = \frac{\pi}{e} - 1 in the above inequality, we get:
e^{\frac{\pi}{e} - 1} > \frac{\pi}{e} \implies e^{\frac{\pi}{e}} > \pi \implies e^{\pi} > \pi^{e}
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