Compare $20^{21}$ and $21^{20}$ using Newton Binomial Expansion.

Question

Compare $20^{21}$ and $21^{20}$ using Newton Binomial Expansion. I mean compare them using $(a+b)^n$ . I think writing 21 = 20 +1 may help but i can't get to an answer.

Answer 1

$$\frac{21^{20}}{20^{20}}=\left(1+\frac{1}{20}\right)^{20} = \sum_{k=0}^{20}\binom{20}{k}\frac{1}{20^k}<\sum_{k=0}^{20}\frac{1}{k!}<1+1+\frac{1}{2}+\frac{18}{2}<12 $$ hence $21^{20} \color{red}{<} 20\cdot 20^{20} = 20^{21}$.

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This is more or less the same argument leading to $$ \lim_{n\to +\infty}\left(1+\frac{1}{n}\right)^n = \sum_{n\geq 0}\frac{1}{n!}.$$