Proving the language $L= \{0^n 1^m \space | \space m \equiv 0 \space mod \space n, \space n \geq 2 \}$ is not regular using the pumping lemma
I am trying to learn about applying the pumping lemma and I'm not really sure how to go about proving this language isn't regular with the pumping lemma:
$L= {0^n 1^m \space | \space m \equiv 0 \space mod \space n, \space n \geq 2 }$
Now I realize that the condition $m \equiv 0 \space mod \space n$, is essentially saying that $m$ is some multiple of $n$. Is it possible that could you go about proving that $L$ is not regular in the way that you can prove that $L = {0^n 1^n}$ is not regular since $m$ is a multiple of $n$, since $m = kn$ (where $k$ is some integer)?
Updated:
-- My attempt at the proof:
If the language $L$ is regular, then by the pumping lemma $\exists \space p \space | \space \forall s \in L \cap \Sigma ^{\geq p} $.
Next by the pumping lemma $\exists x, y, z : s = xyz$, where $s$ is a string such that:
(1) $|y|\geq1$
(2) $|xy|\leq p$
(3) $\forall i\geq0, xy^iz \in L$
Now suppose we let $m = kp$, where $k$ is some integer, let the string $s = 0^p 1^{kp}, s\in L$ and $|s|\geq p$. By the pumping lemma the decomposition of $s$ is defined by $s = xyz$. Now we show that $\forall x, y, z$ that (1)-(3) do not hold.
If (1) and (2) hold, then $s=0^p1^{kp} = xyz$, with $|xy|\leq p$ and $|y|\geq 1$.
So $x = 0^u, y=0^v, z=0^w1^{kp}$
$u+v \leq p$, $v \geq 1 $, $w \geq 0$
$u+v+w = kp$
But (3) fails for $i=2$ since $xy^2z = 0^u0^v0^v0^w1^{kp} = 0^{u+2v+w}1^{kp} \not\in L$ since $u+2v+w \neq kp $
Hence $L$ is not a regular language.
Is this the correct way to go about this proof? 875