Prove ${x \choose k}1/x^k < {y \choose k}1/y^k$ given that $x < y$ and $x, y \in \mathbb{N}$

Question

I am struggling with this proof because of the $1 / x^k$ and $1 / y^k$ part. Proving the inequality of the combinations just boils down to axioms, but I'm not sure how to prove the entire statement. I tried assuming that $y = x + a$ for some $a \in \mathbb{N}$ and then using induction for $k+1$, but I don't know if that's the correct way to do it.

Answer 1

Rewrite the inequality as $$ \frac{y^kx(x-1)\ldots(x-k+1)}{k!} < \frac{x^ky(y-1)\ldots(y-k+1)}{k!}. $$ Cancelling $k!$ on both sides and rearranging gives $$ \left(1-\frac{1}{x}\right)\ldots\left(1-(\frac{k-1}{x})\right) < \left(1-\frac{1}{y}\right)\ldots\left(1-(\frac{k-1}{y})\right), $$ which holds because $x < y$.

(I assumed that $k$ is a natural number that is not more than $x$ or $y$.)