Using binomial theorem, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$
Now, consider ${k\choose r}\frac{1}{k^r}=\frac{k!}{r!(k-r)!}\frac{1}{k^r}=\frac{(1)(1-\frac{1}{k})(1-\frac{2}{k})...(1-\frac{r-1}{k})}{r!}\lt \frac{1}{r!}$
Thus, $(1+\frac{1}{k})^k=1+{k\choose 1}\frac{1}{k}+{k\choose 2}\frac{1}{k^2}+\cdots+{k\choose r}\frac{1}{k^r}+\cdots+{k\choose k}\frac{1}{k^k}$ $ \lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}$ $\lt 1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{r!}+\cdots+\frac{1}{k!}+\frac{1}{(k+1)!}+\cdots =e$
Hence $(1+\frac{1}{k})^k\lt e\lt 3$