Does n^n or a Complex Product Formula Grow Faster Asymptotically?

[Dragonfall]

$$n^n$$

or

$$\prod_{r=0}^{n-1}\left(\begin{array}{c}\sum_{j\leq r}\left(\begin{array}{c}n\\ j\end{array}\right)\\2^r\end{array}\right)$$

Asymptotically, I mean.

 

[CRGreathouse]

Since
$$\prod_{r=0}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}={0\choose1}\prod_{r=1}^{n-1}{{\sum_{j<r}{n\choose j}}\choose{2^r}}=0$$
I'd have to go with $n^n$.

 

[Dragonfall]

How about now?

 

[CRGreathouse]

It seems that the product grows faster.