N Choose k is not greater than n^k

Theorem

 * $\displaystyle {n \choose k} < n^k$

for all $n,k \in \N$, $1 < k \le n$; here $\displaystyle {n \choose k}$ is $n$ choose $k$.

Proof 1
Note that for $k>1$, the product considered has at least two factors, and hence at least one factor which is strictly less that $n$.

Alternative proof
Let $N$ be the set $\{1,\ldots, n\}$, and let $K= \{1,\ldots, k\}$. Then ${n \choose k}$ is the number of strictly increasing functions from $K$ to $N$, whereas $n^k$ is the number of all functions from $K$ to $N$. Now if $K$ has more than one element, then not all functions from $K$ to $N$ are strictly increasing; hence a strict inequality holds.