N Choose k is not greater than n^k

Theorem

 * $\displaystyle \forall n, k \in \N, 1 < k \le n: {n \choose k} < n^k$

where $\displaystyle {n \choose k}$ is a binomial coefficient.

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 $n$.

Proof 2
Let:
 * $N = \left\{{1, \ldots, n}\right\}$
 * $K = \left\{{1, \ldots, k}\right\}$

From Cardinality of Set of Strictly Increasing Mappings, $\displaystyle \binom n k$ is the number of strictly increasing mappings from $K$ to $N$.

From Cardinality of Set of All Mappings, $n^k$ is the number of all mappings from $K$ to $N$.

We have that $K$ has more than one element, and so therefore does $n$.

The mapping $f: K \to N: \forall s \in K: f \left({s}\right) = 1$ is clearly not a strictly increasing mapping.

So not all mappings from $K$ to $N$ are strictly increasing.

Hence a strict inequality holds, and so $\displaystyle {n \choose k} < n^k$.