Prove the statement by the Principle of Mathematical Induction: For any natural number $n$,$x^{n}-y^{n}$ is divisible by $x-y$,where $x$ and $y$ are any integers with $x \neq y$.

(N/A) Let $P(n): x^{n}-y^{n}$ is divisible by $(x-y)$ for all $n \in \mathbb{N}$.
Step $1$: For $n=1$,$P(1): x^{1}-y^{1} = x-y$,which is clearly divisible by $(x-y)$. Thus,$P(1)$ is true.
Step $2$: Assume $P(k)$ is true for some $k \in \mathbb{N}$,i.e.,$x^{k}-y^{k} = m(x-y)$ for some integer $m$. (Equation $i$)
Step $3$: For $n=k+1$,we need to show $P(k+1): x^{k+1}-y^{k+1}$ is divisible by $(x-y)$.
$x^{k+1}-y^{k+1} = x^{k+1} - x^{k}y + x^{k}y - y^{k+1}$
$= x^{k}(x-y) + y(x^{k}-y^{k})$
Substituting from Equation $i$:
$= x^{k}(x-y) + y(m(x-y))$
$= (x-y)(x^{k} + my)$
Since $(x^{k} + my)$ is an integer,$x^{k+1}-y^{k+1}$ is divisible by $(x-y)$.
Thus,$P(k+1)$ is true whenever $P(k)$ is true.
By the Principle of Mathematical Induction,$P(n)$ is true for all $n \in \mathbb{N}$.