sequences and series - Prove that $n^n$ is greater than $1cdot3cdot5cdots(2n-1)$

Prove that $\ n^n \ge (1)(3)(5)\cdots(2n-1)$

I can't think of how to start answering this question and it would be great help if someone could explain how I should go about doing it.

Note:It is a sequence and series question(AP,GP,HP)

Answer

Only noticed this question today. Although the selected answer is quite nice and arguably simpler than the argument below, none of the posted answers address what appeared to be the original intent of establishing the inequality using the Arithmetic Mean-Geometric Mean Inequality. For this, simply notice that
$$ 1+3+\ldots+(2n-1)=n^2, $$
which can be easily proved, for instance, using induction. By the AM-GM inequality, it follows that
$$n=\frac{1+3+\ldots+(2n-1)}n\ge\sqrt[n]{1\cdot 3\cdot\ldots\cdot (2n-1)}, $$
and we are done.