I have this formula which seems to work for the product of the first n odd numbers (I have tested it for all numbers from $1$ to $100$):
$$\prod_{i = 1}^{n} (2i - 1) = \frac{(2n)!}{2^{n} n!}$$
How can I prove that it holds (or find a counter-example)?
The idea is to "complete the factorials":
$$ 1\cdot 3 \cdot 5 \cdots (2n-1) = \frac{ 1 \cdot 2 \cdot 3 \cdot 4 \cdots (2n-1)\cdot (2n) }{2\cdot 4 \cdot 6 \cdots (2n)} $$
Now take out the factor of $2$ from each term in the denominator:
$$ = \frac{ (2n)! }{2^n \left( 1\cdot 2 \cdot 3 \cdots n \right)} = \frac{(2n)!}{2^n n!}$$
A mathematician may object that there is a small gray area about what exactly happens between those ellipses, so for a completely rigorous proof one would take my post and incorporate it into a proof by induction.
For the induction argument, $$\begin{align*} \prod_{i=1}^{n+1}(2i-1)&=\left(\prod_{i=1}^n(2i-1)\right)(2n+1)\\ &= \frac{(2n)!(2n+1)}{2^n n!} \end{align*}$$ by the induction hypothesis. Now multiply that last fraction by a carefully chosen expression of the form $\dfrac{a}a$ to get the desired result.
$$\Pi_{k=1}^n(2k-1) = \Pi_{k=1}^n(2k-1) \frac{\Pi_{k=1}^n(2k)}{\Pi_{k=1}^n(2k)} =\frac{\Pi_{k=1}^{2n}k}{2^n\Pi_{k=1}^n k} = \frac{(2n)!}{2^n n!}$$
