How is this summation expression transformed?

Question

I am solving one math problem. I could not understand the following transformation described below. I am guessing the denominator is re-written as a factorial of 2K and unnecessary even terms are canceled by the numerator. But, I have no idea how the transformation is done. I would appreciate if you could clarify the transformation step by step

$$ = \sum_{k=0}^{\infty } \frac{1\times 3\times ...\times (2k-1) }{2^{k}(k!)}x^{k} =\sum_{k=0}^{\infty }\frac{(2k)!}{2^{2k}\times (k!)^2}x^{k}$$

Answer 1

Hint

$$(2k-1)!!=\frac{(2k)!}{2^kk!} $$ where $(2k-1)!!$ denotes the double factorial.