Formula for $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots$ I was doing an exercise about the integral $$ I_{n} = \int_{0}^{1} (1-x^{2})^{n}\, \mathrm{d}x $$ and I tried two approaches. On the one hand, by substituting $x = \sin u $ I found that $$ I_{n} = \frac{2}{3} × \frac{4}{5} × \cdots × \frac{2n}{2n+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right).$$
On the other hand, the binomial expansion formula gave me $$I_{n} = \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1}.$$
So I find that $$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right),$$
but I want to understand this identity without integration. Is there a direct way to prove this? Perhaps by induction or generating polynomials or something.
I tried induction, and found that proving the identity is equivalent to proving $$ \sum_{k=0}^{n} \binom{n}{k}(-1)^{k} \left( \frac{1}{2n+3} - \frac{1}{2k+3} \right) = 0 $$
but here I get stuck.

Inductive proof.
Letting $$f(n)=\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\ldots\left(1-\frac{1}{2n+1}\right)$$
and $$g(n)=\sum_{k=0}^n\binom{n}{k}\frac{(-1)^k}{2k+1}$$
our goal is to prove that $f (n) = g (n) $ for all $n $.
The key to the induction is that we also have:

Lemma: For $n>0$:
  $$g(n)=\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k+1}2k}{2k+1}$$
This is because $\frac{-2k}{2k+1}=\frac{1}{2k+1}-1$ and $\sum (-1)^k\binom{n}{k}=0$.

We'll also use that $n\binom{n-1}{k}=(n-k)\binom{n}{k}$.
Induction step:
Assume:
$$g(n-1)= f(n-1)$$
Then:
$$\begin{align}(2n+1)f(n)&=2nf(n-1)\\&=2ng(n-1)\\&=2n\sum_{k=0}^{n-1}\binom{n-1}{k}\frac{(-1)^{k}}{2k+1}\\
&=2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^k(n-k)}{2k+1}\\
&=2n\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^k}{2k+1}+2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{k+1}k}{2k+1}\\
&=2n\left(g(n)-\binom{n}{n}\frac{(-1)^n}{2n+1}\right)+2\sum_{k=0}^{n-1}\binom{n}{k}\frac{(-1)^{k+1}k}{2k+1}\\
&=2ng(n)+\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k+1}2k}{2k+1}\\&=(2n+1)g(n)
\end{align}$$
