Difference of two variables raised to the $k$. Let $k \ge 2$ be an integer and let $0 < y < x$ where $x, y$ are integers. Show that $x^k-y^k > 2$.
I believe I have proved this statement but want to make sure it is sound.
Proof: Since $0 < y < x$ and $x$ and $y$ are integers, $x \ge 2$ and $y \ge 1$ and $x^k>y^k$. So $x^k-y^k > 0$. Thus $0 < 2^k-1^k \le x^k-y^k.$ Since $k$ is at least $2$, $3 = 2^2-1^2 \le 2^k-1^k \le x^k-y^k $. Hence, $x^k-y^k > 2$.

The only part of your proof that I find is not fully explained is why $2^k-1^k\leq x^k-y^k$.

Another way to see it is that
$$
x^k-y^k=(x-y)(x^{k-1}+x^{k-2}y+...+x y^{k-2}+y^{k-1})
$$
where $(x-y)\geq 1$ and $(x^{k-1}+x^{k-2}y+...+x y^{k-2}+y^{k-1})$ consists of $k$ terms all greater than or equal to $1$.
