Route to go:
First show that $P\left(1\right)$ and $P\left(2\right)$are
both true.
Setting $B:=A_{1}\cup\cdots\cup A_{k}$ by applying $P\left(2\right)$
we find:
$\tag1 N\left(A_{1}\cup\cdots\cup A_{k}\cup A_{k+1}\right)=N\left(B\cup A_{k+1}\right)=N\left(B\right)+N\left(A_{k+1}\right)-N\left(B\cap A_{k+1}\right)$
Under assumption that $P(k)$ is true find expressions for: $$N\left(B\right)=N\left(A_{1}\cup\cdots\cup A_{k}\right)$$
and for $$N\left(B\cap A_{k+1}\right)=N\left(\left(A_{1}\cap A_{k+1}\right)\cup\cdots\cup\left(A_{k}\cap A_{k+1}\right)\right)$$
Substitute these expressions in (1).
edit:
$N\left(B\right)=\sum_{i=1}^{k}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\right)$
$N\left(B\cap A_{k+1}\right)=\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)-\cdots+\left(-1\right)^{k+1}N\left(\left(A_{1}\cap A_{k+1}\right)\cap\cdots\cap\left(A_{k}\cap A_{k+1}\right)\right)=\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)-\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
Substitution of this in the (1) gives the following RHS:
$\sum_{i=1}^{k}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+1}N\left(A_{1}\cap\cdots\cap A_{k}\right)+N\left(A_{k+1}\right)-\sum_{i=1}^{k}N\left(A_{i}\cap A_{k+1}\right)+\cdots+\left(-1\right)^{k+2}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
After a rearrangement of the terms we find:
$N\left(A_{1}\cup\cdots\cup A_{k}\cup A_{k+1}\right)=\sum_{i=1}^{k+1}N\left(A_{i}\right)-\sum_{1\leq i<j\leq k+1}N\left(A_{i}\cap A_{j}\right)+\cdots+\left(-1\right)^{k+2}N\left(A_{1}\cap\cdots\cap A_{k}\cap A_{k+1}\right)$
wich is exactly statement $P\left(k+1\right)$.
The irrationality of $\sqrt 2^{\sqrt 2}$ (in fact, its transcendence) follows immediately from the Gelfond Schneider Theorem. This was the issue that motivated Hilbert's $7^{th}$ Problem.
The beauty of the argument here is its simplicity. It's a perfectly valid, non-constructive, argument. The claim is demonstrated though no explicit example is produced.
Here is a simple, constructive, way to settle the original issue: $$\sqrt 3^{\log_34}=2$$
Of course, $\sqrt 3$ is irrational.
To see that $\log_3 4$ is irrational work by contradiction. $$\log_3 4=\frac ab\implies 4=3^{\frac ab}\implies 4^b=3^a$$ But if $a,b\in \mathbb N$ then this contradicts unique factorization.
Best Answer
Your factorisation is incorrect. Use $x^{n+1}-y^{n+1}=x(x^n-y^n)+y^n(x-y)$.