N Choose k is not greater than n^k/Proof 2

Theorem

$\forall n \in \Z, k \in \Z: 1 \le k \le n: \dbinom n k < n^k$

where $\dbinom n k$ is a binomial coefficient.

Equality holds when $k = 0$ and $k = 1$.

We dismiss the cases where $k < 0$ by observing that in such cases $\dbinom n k = 0$ while $n^k > 0$.

Similarly we dismiss $k = 0$: we have $\dbinom n 0 = 1 = n^0$.

Let:

$N = \set {1, \ldots, n}$

$K = \set {1, \ldots, k}$

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

For $k = 1$ there is exactly 1 mapping from $K$ to $N$ which is trivially strictly increasing.

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

The mapping $f: K \to N: \forall s \in K: \map f s = 1$ 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 $\dbinom n k < n^k$.

$\blacksquare$