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**o_O** Assume it is true for $\displaystyle n = k$, i.e. $\displaystyle 1 \cdot 3 \cdot 5 \cdots (2k-1) = \frac{(2k-1)!}{2^{k-1}(k-1)!}$

It remains to show that it is also true for $\displaystyle n=k+1$, i.e. $\displaystyle 1 \cdot 3 \cdot 5 \cdots (2k-1)(2k+1) = \frac{(2k+1)!}{2^{k} \cdot k!}$

Looking at the LHS:

$\displaystyle \begin{aligned} \underbrace{1 \cdot 3 \cdot 5 \cdots (2k-1)}_{\text{Our assumption}}(2k+1) & = \frac{(2k-1)!}{2^{k-1}(k-1)!} \cdot (2k+1) \\ & = \frac{(2k+1)(2k-1)!}{2^{k-1}(k-1)!} \cdot {\color{red} \frac{2k}{2k}} \\ & = \frac{(2k+1){\color{red}2k}(2k-1)!}{{\color{red}2}\cdot 2^{k-1} \cdot {\color{red}k}(k-1)!}\end{aligned}$

Can you change the numerator into a single factorial? Simplify the denominator as well. Then you should be good.