By am- gm inequality for m odd numbers gives required inequality I.e. Given inequality >= 1

by AM-GM we have ${(1.3.5........(2m-1))}^{1/m}\leq \frac{1+3+5+....+(2m-1)}{m}=m\Longrightarrow \dfrac{m^m}{1\cdot 3\cdot 5\cdot \ldots \cdot(2m-1)}\ge 1$