More on Wallis' product

Here's a thought. Suppose we knew Wallis' product
$$\frac{2}{1}\frac{2}{3}\frac{4}{3}\frac{4}{5}\frac{6}{5}\frac{6}{7}\cdots =\frac{\pi}{2}.$$ Define $I_{2n}=\tfrac{1}{2}\tfrac{3}{4}\cdots \frac{2n-1}{2n}\frac{\pi}{2}$ and $I_{2n+1}=\tfrac{2}{3}\tfrac{4}{5}\cdots \frac{2n}{2n+1}$.
If we truncate the left hand side of Wallis' product after $2n$ terms we make it less, and this implies $I_{2n+1}/I_{2n}<1$. Now we can complete the proof of Stirling's formula without ever needing to consider $I_n=\int_0^{\pi/2}\sin^n\theta\,d\theta$.

Of course a standard way to prove Wallis' product is to work with $I_n=\int_0^{\pi/2}\sin^n\theta\,d\theta=(n-1)(I_{n-2}-I_n)$, but there are other ways to prove it that require just basic algebra, Pythagoras' theorem and the formula for the area of a circle. See here.