$$c_n=\left(1+\frac{1}{n}\right)^n$$
Binomio de Newton:
$$c_n=\sum_{i=0}^n C_i^n \left(\frac{1}{n}\right)^i (1)^{n-i}=\sum_{i=0}^n C_i^n \left(\frac{1}{n}\right)^i$$
$$C_i^n=\frac{n!}{i!(n-i)!}$$
Entonces
$$c_n=\sum_{i=0}^n \frac{n!}{i!(n-i)!} \left(\frac{1}{n}\right)^i$$
$$c_n=\sum_{i=0}^n \frac{n!}{i!(n-i)!n^i}$$
$$\frac{n!}{i!(n-i)!n^i}=\frac{\prod_{j=n-i+1}^n j}{i!n^i}=\frac{\prod_{j=n-i+1}^n \frac{j}{n}}{i!}$$
Como $$\frac{j}{n}\leq 1 \forall j\in \lbrace n-i+1, ..., n\rbrace$$, entonces
$$\frac{n!}{i!(n-i)!n^i}\leq \frac{\prod_{j=n-i+1}^n 1}{i!}=\frac{1}{i!} \forall i\in \lbrace 0, ..., n\rbrace$$
Entonces
$$c_n=\sum_{i=0}^n \frac{n!}{i!(n-i)!n^i}\leq \sum_{i=0}^n \frac{1}{i!}$$
$$c_n \leq a_n$$