Denote the exponential generating function $f:\mathbb{N}\to K$ by $E_f(x)$, where $K$ stands for a field of characteristic $0$ such as $\mathbb{C}$ with some indeterminates adjoined)
Let $f(n)=1\times3\times\cdots\times(2n-1),g(n)=2^n\times n!$.
Can someone show me how can $E_g(x)$ be identical to $E_f(x)^2$? In another words, how is $E_g(x)=e_f(x)^2$?
Help appreciated!
Well, let's put it in this way,
$$\because1\times3\times\cdots\times(2n-1)=\dfrac{2n!}{2^n\times n!}$$ \begin{align*} \therefore E_f(x)&=\sum_{n\geq0}\dfrac{x^n}{2^n}{2n\choose n}\\ \\ &=\frac1{\sqrt{-2x+1}} \end{align*} \begin{align*} \therefore E_f(x)^2&=\frac1{-2x+1} \\ \\ &=\sum_{n\geq0}2^n\times x^n \\ \\ &=\sum_{n\geq0}(2^n\times n!)\times\frac{x^n}{n!} \\ \\ &=E_g(x)\ \ \square \end{align*}