I heard that the following problem is for high school students.

Prove that $e^\pi > 21$.

Without a proof, you can use the two facts: $e > 2.71$ and $\pi > 3.14$.

My solution is the following:

$2.7^4 = 53.1441 > 53$.

$2.7^8 > 53^2 = 2809$.

$e^{10} > 2.7^{10} > 2.7^2 \times 2809 = 7.29 \times 2809 = 20477.61 > 19683 = 3^9$.

$10 = \log(e^{10}) > \log(3^9) = 9\times \log 3.$

$\frac{10}{9} > \log 3$.

$e^2 > 2.7^2 = 7.29 > 7$.

$2 = \log(e^2) > \log 7$.

$\pi > 3.14 > 3.11\cdots = \frac{10}{9} + 2 > \log 3 + \log 7 = \log 21$.

So, $e^\pi > 21$.

Honestly speaking, I used Google Sheets to solve this problem.

Please tell me an easier solution.