Can we let $n$ go to $\infty$ after applying the fundamental theorem of algebra to a polynomial of degree $n$?

Recently I came across a proof of the infinite product for $$\sin z$$ (https://www.sciencedirect.com/science/article/pii/0022247X77902347). It applies the fundamental theorem of algebra to $$p_{n}(z)=\dfrac{1}{2}\left(\left(1+\frac{z}{n}\right)^{n}-\left(1-\frac{z}{n}\right)^{n}\right),$$ which it factors to a product. But it also states that $$n=2m+1$$ and lets $$n$$ and $$m$$ go to infinity.

Then how is the factorization to a product, by applying the fundamental theorem of algebra, possible?

I'm confused because the fundamental theorem of algebra holds only for finite polynomials, but $$p_{n}(z)$$ is not a finite polynomial if we let $$n$$ go to $$\infty$$.

• It is a standard fact that $a^n-b^n=(a-b)(a^{n-1}+\cdots+b^{n-1}) = \prod_{k=0}^{n-1} (a - z^k b)$ where $z:=e^{2\pi i/n}$ is a primitive $n$-th root of unity. – Somos May 29 '19 at 1:13
• I'm familiar with roots of unity but can't find a proof of $(a-b)(a^{n-1}+\cdots +b^{n-1})=\prod_{k=0}^{n-1}\left(a-be^{\frac{2k\pi i}{n}}\right).$ – Abo Rakan May 29 '19 at 12:06
• Even if it is proved, how can we let $n$ go to $\infty$? Using the roots of unity, we assumed finite $n$, didn't we? – Abo Rakan May 29 '19 at 12:19
• They are both equal to $a^n-b^n$. In limit proofs we always let $n\to\infty$. For example, $2^{-n}\to0$ as $n\to\infty$ where $n$ is always finite. – Somos May 29 '19 at 12:22

I suspect that if you read that paper carefully you would find that Eberlein's purpose is (in part) to prove rigorously that Euler's seemingly casual "letting $$n$$ go to infinity" produces mathematics that's correct by 20th century standards.