Des factorielles utiles!

On remarque que
$$2\times4\times6\times\cdots\times(2n) = (2\times1) \times( 2\times2) \times \cdots \times (2\times n).$$En factorisant les puissances de $2$, on obtient
$$2\times4\times6\times\cdots\times(2n) = 2^n(1\times2\times\cdots\times n) = 2^{n} n!.$$

Notons
$$P=1\times3\times5\times\cdots\times(2n+1).$$Comme
$$1=\frac{2\times4\times6\times\cdots\times(2n)} {2\times4\times6\times\cdots\times(2n)},$$on obtient
$$P = 1\times3\times5\times\cdots\times(2n+1) \times \frac{2\times4\times6\times\cdots\times(2n)} {2\times4\times6\times\cdots\times(2n)}.$$Ainsi,
$$P = \frac{ 1\times2\times3\times4\times\cdots\times(2n)\times(2n+1) } {2\times4\times6\times\cdots\times(2n)}.$$Or
$$1\times2\times3\times4\times\cdots\times(2n)\times(2n+1) = (2n+1)!$$et, d’après la question précédente,
$$2\times4\times6\times\cdots\times(2n) = 2^n n!.$$Par conséquent,
$$P = \frac{(2n+1)!}{2^n n!}.$$