Binomial inequality : $\binom{n}{k} \leq n^k$

Question

$\forall n \in \mathbb N$ and $k \in [[0,n]]$, show that :

$$\binom{n}{k} \leq n^k$$

I already showed that : $\frac{1}{n^k}\binom{n}{k}$$\leq$$\frac{1}{(n+1)^k}\binom{n+1}{k}$

Answer 1

Simply note that $$\binom{n}{k}=\frac{n(n-1)\cdots(n-k+1)}{k!}\leq \underbrace{n(n-1)\cdots(n-k+1)}_{\text{$k$ factors $\leq n$}}\leq n^k.$$ Actually, the same approach leads to the following stronger inequality $$\binom{n}{k}\leq \frac{n^k}{k!}.$$

Answer 2

HINT: note that

$$\binom{n}{k}=\frac{n!}{k!(n-k)!}=\frac{n(n-1)\cdots(n-k+1)}{k!}$$

Answer 3

Note that$$\binom nk=\frac{n(n-1)\ldots(n-k+1)}{k!}=\frac nk\times\frac{n-1}{k-1}\times\cdots\times(n-l+1)\leqslant n^k,$$since each of the $k$ factors is smaller than or equal to $n$.