This (reputable) Youtube video claims to have found a new proof for Wallis' Product $$\prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot \frac{2n}{2n+1}\right) = \frac{\pi}{2}.$$
There is a supplementary blog post located here that provides some detail and supplemental rigour.
The general gist of the argument is that we consider the $n$th roots of unity, along with two points on the unit circle. We then take a product of their distances to the roots of unity and use Vieta's formula to get a closed form for the products.
Is this a genuinely new proof for Wallis' Product? Is it completely rigorous? If not, do you know of any existing literature that outlines a similar proof?
I personally enjoyed their exhibition of the proof and think it's great that the channel is able to push maths into a more mainstream audience, I'm mainly asking this out of curiosity. Feel free to add/replace the tags with more appropriate ones.
Mostly no.
Using this sort of geometric idea to prove identities appears in this paper by Wästlund, where a similar construction is used to prove $\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$, and more generally $$ \pi^2\csc^2{\pi x} = \sum_{n=-\infty}^{\infty} \frac{1}{(x-n)^2}. $$ On the other hand, Nick Lord (Math. Gaz. 100, No. 549, pp. 429-434 (2016)) suggests that the algebraic version of this proof goes back to Cauchy, at least. We will see that this proof is analogous.
Having skimmed the video (the author apparently not having the good grace to write the proof down), it seems pretty clear that the proof is a geometric analogue of the following elementary proof by Yaglom and Yaglom in 1953 (A. M. Yaglom, I. M. Yaglom, “An elementary derivation of the formulas of Wallis, Leibnitz [sic] and Euler for the number $\pi$”, Uspekhi Mat. Nauk, 8:5(57) (1953), pp. 181–187. Available here. The same proof is detailed in their book "Challenging Mathematical Problems with Elementary Solutions", Problem 141 ff.)
We can prove by elementary polynomial algebra that $$ \sin^2{\frac{2\pi}{4m}}\sin^2{\frac{4\pi}{4m}}\sin^2{\frac{6\pi}{4m}} \dotsm \sin^2{\frac{(2m-2)\pi}{4m}} = \frac{m}{2^{2m-2}} \\ \sin^2{\frac{\pi}{4m}}\sin^2{\frac{3\pi}{4m}}\sin^2{\frac{5\pi}{4m}} \dotsm \sin^2{\frac{(2m-1)\pi}{4m}} = \frac{1}{2^{2m-1}} $$ (this is done in the paper, but is fairly well-known). We then use these to show that $$ \frac{\sin^2{\frac{2\pi}{4m}}}{\sin^2{\frac{\pi}{4m}}} \frac{\sin^2{\frac{2\pi}{4m}}}{\sin^2{\frac{3\pi}{4m}}} \frac{\sin^2{\frac{4\pi}{4m}}}{\sin^2{\frac{3\pi}{4m}}} \frac{\sin^2{\frac{4\pi}{4m}}}{\sin^2{\frac{5\pi}{4m}}} \dotsm \frac{\sin^2{\frac{(2m-2)\pi}{4m}}}{\sin^2{\frac{(2m-3)\pi}{4m}}} \frac{\sin^2{\frac{(2m-2)\pi}{4m}}}{\sin^2{\frac{(2m-1)\pi}{4m}}} = m\sin{\frac{\pi}{2m}} $$ and $$ \frac{\sin^2{\frac{2\pi}{4m}}}{\sin^2{\frac{3\pi}{4m}}} \frac{\sin^2{\frac{4\pi}{4m}}}{\sin^2{\frac{3\pi}{4m}}} \frac{\sin^2{\frac{4\pi}{4m}}}{\sin^2{\frac{5\pi}{4m}}} \frac{\sin^2{\frac{6\pi}{4m}}}{\sin^2{\frac{5\pi}{4m}}} \dotsm \frac{\sin^2{\frac{(2m-2)\pi}{4m}}}{\sin^2{\frac{(2m-1)\pi}{4m}}} \frac{\sin^2{\frac{2m\pi}{4m}}}{\sin^2{\frac{(2m-1)\pi}{4m}}} = m\tan{\frac{\pi}{4m}}. $$ Obviously what happens now is a sandwiching argument: we have the trigonometric identity $$ \frac{\sin{(k-1)a}}{\sin{ka}}\frac{\sin{(k+1)a}}{\sin{ka}} = 1-\left( \frac{\sin{a}}{\sin{ka}} \right)^2, $$ and the corresponding algebraic identity $$ \frac{(k-1)a}{ka} \frac{(k+1)a}{ka} = 1- \left( \frac{a}{ka} \right)^2. $$ Lastly, $\sin{x}/x$ is decreasing for $0<x<\pi/2$, so in this range $$ \frac{\sin{a}}{\sin{ka}} > \frac{a}{ka}, $$ and so $$ \frac{\sin{(k-1)a}}{\sin{ka}}\frac{\sin{(k+1)a}}{\sin{ka}} < \frac{(k-1)a}{ka} \frac{(k+1)a}{ka}. $$ This, combined with the big products, gives us the two inequalities $$ \frac{\frac{2\pi}{4m}}{\frac{\pi}{4m}} \frac{\frac{2\pi}{4m}}{\frac{3\pi}{4m}} \frac{\frac{4\pi}{4m}}{\frac{3\pi}{4m}} \frac{\frac{4\pi}{4m}}{\frac{5\pi}{4m}} \dotsm \frac{\frac{(2m-2)\pi}{4m}}{\frac{(2m-3)\pi}{4m}} \frac{\frac{(2m-2)\pi}{4m}}{\frac{(2m-1)\pi}{4m}} < m\sin{\frac{\pi}{2m}} \\ \frac{\frac{2\pi}{4m}}{\frac{3\pi}{4m}} \frac{\frac{4\pi}{4m}}{\frac{3\pi}{4m}} \frac{\frac{4\pi}{4m}}{\frac{5\pi}{4m}} \frac{\frac{6\pi}{4m}}{\frac{5\pi}{4m}} \dotsm \frac{\frac{(2m-2)\pi}{4m}}{\frac{(2m-1)\pi}{4m}} \frac{\frac{2m\pi}{4m}}{\frac{(2m-1)\pi}{4m}} > m\tan{\frac{\pi}{4m}}. $$ Cancelling and rearranging, this gives $$ \frac{\pi}{2} \frac{\sin{\frac{\pi}{2m}}}{\frac{\pi}{2m}} > \frac{2}{1} \frac{2}{3} \frac{4}{3} \frac{4}{5} \dotsm \frac{2m-2}{2m-3} \frac{2m-2}{2m-1} > \frac{\pi}{2} \left( 1 - \frac{1}{2m} \right) \frac{\tan{\frac{\pi}{4m}}}{\frac{\pi}{4m}}, $$ and sending $m \to \infty$ gives the result.
The proof in the video therefore appears to be a Wästlund-esque geometric presentation of this proof, skipping over a lot of the rigour (in particular, the lengths of the chords are obviously closely related to the $\sin{(k\pi/4m)}$ in the Yagloms' proof). Indeed, this makes perfect sense, given that the Youtube channel also has a video about Wästlund's construction.
(Also of interest: Wästlund has another, different geometric proof of the Wallis product, using a subdivision of the area of the circle, in the American Mathematical Monthly (AMM 114, pp. 914–917 (2007)), available here.)