Which is larger??

From AM-GM inequality , we have:

$\begin{aligned} \frac {a_1+a_2+a_3+\cdots + a_n}n & \ge \sqrt[n]{a_1a_2a_3\cdots a_n} & \small \blue{\text{For different }a_k} \\ \frac {1+3+5+\cdots+(2n-1)}n & > \sqrt[n]{1 \times 3 \times 5 \times \cdots \times (2n-1)} \\ \frac {\frac n2 (1+(2n-1))}n & > \sqrt[n]B \\ n & > \sqrt[n]B & \small \blue{\text{Raise both sides to the power of }n} \\ n^n & > B \end{aligned}$

Therefore, $\boxed{A > B}$ .

The arithmetic mean of the terms of the sequence-1,3,5,7,...,2n-1= $\frac{n^2}{n}=n$

And the geometric mean is $(1*3*5*...*(2n-1))^{\frac{1}{n}}$

So, n>(1 3 5*...(2n-1)) $^{\frac{1}{n}}$

So, $n^n>(1*3*5*...(2n-1))$