Factorial usage within proof using the pumping lemma
Let's choose a different word in step 2, say $s' = 0^a1^b$, where $a$ and $b$ depend on $P$ and $a \geq P$ (to force an $y$ that has the form $y=0^m$).

It's important to show that for any length $m, 1\leq m \leq P$ there is no valid division $s' = xyz, y = 0^m$ (valid means it fulfils the conditions of the pumping lemma).

So for every such $m$, we need an $i$ such that:
$xy^iz = x0^{im}z = 0^{a-m+im}1^b  = 0^{a+(i-1)m}1^b {! \atop =}  0^b1^b$.
This means that $m$ must divide $b-a$ for every $m$. One number $b-a$ that fulfils this, is $P!$, so e.g. $a=P$, $b=P+P!$, the least common multiple of $1,\dots,m$ is another possible choice for $b-a$.