Analytical proof required $\binom{n}{r} < (n+1)^r$

Question

$\binom{n}{r} < (n+1)^r$

I need help to prove the above expression using some analytical method.

Answer 1

$$\binom{n}{r}=\frac{n!}{r!(n-r)!}<\frac{n!}{(n-r)!}=\underbrace{n(n-1)\dots(n-r+1)}_{r-terms}<\underbrace{(n+1)(n+1)\dots(n+1)}_{r-terms}=(n+1)^r$$

Answer 2

Hint:

Equality holds at $r = 0$, so inequality starts at $r = 1.$

Simplifying assumption made that $2 \leq n \in \mathbb{Z^+}$ and that $r \in \{1,2,\cdots, n\}.$

With base case of $r=1$ established, do induction on $r$, noting the scaling factor on each side of the inequality.

Answer 3

$$ \binom{n}{r} = \frac{n!}{r!(n-r)!} = \frac{\prod\limits_{k=1}^{n}k}{r!\prod\limits_{k=1}^{n-r}k} = \frac{\prod\limits_{k=n-r+1}^{n}k}{r!} \le \frac{\prod\limits_{k=n-r+1}^{n}n}{r!} = \frac{n^r}{r!} \le n^r \le \sum\limits_{k=0}^{r} \binom{r}{k} n^k = (n+1)^r $$