Prove $(1 + 1/n)^n \le n+1$

Question

Using principle of mathematical induction, prove that Prove the property coined $\mathcal P (n)$ $$(1 + 1/n)^n \le n+1$$ When $n =1$, $LS =2$, $RS =2$

Hence $\mathcal P (1)$ is true. Let $\mathcal P (k)$ be true Then I have to prove $$(1 + 1/k)^k \le k+1$$

#stuck from here onwards. Cannot figure out a way to bring $k+1$... tried multiplying $k+1$ on both sides and obviously failed

Answer 1

So, the base case is indeed true. Now, assume

$${\left(1+\frac{1}{n}\right)^{n}\leq n+1}$$

Then clearly

$${\Rightarrow \left(1+\frac{1}{n+1}\right)^{n}\leq n+1}$$

(because ${\frac{1}{n+1} < \frac{1}{n}}$). Now from our assumption,

$${\left(1+\frac{1}{n+1}\right)^{n+1}=\left(1+\frac{1}{n+1}\right)\left(1+\frac{1}{n+1}\right)^{n}\leq \left(1+\frac{1}{n+1}\right)(n+1)=n+2}$$

And so by the principle of Mathematical induction, ${\forall\ n \in \mathbb{N}}$ we have

$${\Rightarrow \left(1+\frac{1}{n}\right)^{n}\leq n+1}$$

Answer 2

A simple way would be to use $\left(1+\frac{1}{n}\right)^n < e$.

Then case $n \ge 2$:

$e < 3 \le n + 1$

Case $n = 1$:

$2 = 2$