Proving formula for product of first n odd numbers

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.