How can I simplify $\prod \limits_{l=1}^{a} \frac{1}{4^a} \cdot 16^l$?

Question

How can I simplify $\prod \limits_{l=1}^{a} \frac{1}{4^a} \cdot 16^l$?

I've tried looking at the terms and finding something in there to conlcude what it might be and also took the $n^{th}$ term of $16^l$ into one fraction but that does rather the opposite of simplification.

Answer 1

$$\prod_{l=1}^{a}\frac{16^l}{4^a} = \frac{1}{4^{a^2}}\prod_{l=1}^{a}16^l = 4^{-a^2} 16^{\sum_{l=1}^{a}l} = 4^{-a^2} 4^{a(a+1)}=\color{red}{4^a}.$$

Answer 2

Following Jack's answer and using $\sum_{k=1}^{n} k = n(n+1)/2$ it can be seen that: \begin{align} \prod_{k=1}^{n} \left\{ \frac{x^{2k}}{y^{n}} \right\} &= \left( y^{-n} \right)^{n} \, \left( \prod_{k=1}^{n} x^{2k} \right) \\ &= y^{-n^{2}} \, x^{2 \, \sum_{k=1}^{n} k } = y^{-n^{2}} \, x^{n^{2} + n } \\ &= x^{n} \, \left( \frac{x}{y}\right)^{n^{2}}. \end{align} Examples:

• $y=x$ $$\prod_{k=1}^{n} \left\{ \frac{x^{2k}}{x^{n}} \right\} = x^{n}$$
• $y = x^{2}$ $$\prod_{k=1}^{n} \left\{ \frac{x^{2k}}{x^{2n}} \right\} = \frac{1}{x^{n(n-1)}}$$