Binom açılımından (1+1n)n≥2 olduğu aşikar;
(1+1n)n≤3 için;
\left(1+\dfrac1n\right)^n=\displaystyle\sum_{i=0}^n\dbinom{n}{i}\dfrac{1}{n^i}=1+1+\displaystyle\sum_{i=2}^n\dbinom{n}{i}\dfrac{1}{n^i}
=2+\displaystyle\sum_{i=2}^n\dfrac{n!}{i!(n-1)!}\dfrac{1}{n^i}=2+\displaystyle\sum_{i=2}^n\dfrac{n(n-1)(n-2)...(n-(i-1))}{n^i}\dfrac{1}{i!}
\le 2+\displaystyle\sum_{i=2}^n\dfrac{1}{i!}\le 2+\displaystyle\sum_{i=2}^n\dfrac1{2^{i-1}}<2+\displaystyle\sum_{i=1}^\infty=3
\Box