Simplify $\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{3-k} = \frac{3^{3-k}}{64}$

Question

How can one derive the following simplification?

$$\left(\frac{1}{4}\right)^{k}\left(\frac{3}{4}\right)^{3-k} = \frac{3^{3-k}}{64}$$

Answer 1

$$\begin{array}{lr}\left(\dfrac{1}{4}\right)^k\left(\dfrac{3}{4}\right)^{3-k}&\text{original}\\=\left(\dfrac{1^k}{4^k}\right)\left(\dfrac{3^{3-k}}{4^{3-k}}\right)&\text{distribute the exponents}\\=\dfrac{1^k\cdot 3^{3-k}}{4^k\cdot 4^{3-k}}&\text{combine fractions}\\=\dfrac{3^{3-k}}{4^k\cdot 4^{3-k}}&\text{simplify the 1}\\=\dfrac{3^{3-k}}{4^{k+3-k}}&\text{combine exponents for same base of 4}\\=\dfrac{3^{3-k}}{4^3}&\text{simplify exponent cancelling }k\\=\dfrac{3^{3-k}}{64}&\text{evaluate }4^3\end{array}$$

Answer 2

$$\left(\frac{1}{4}\right)^k\left(\frac{3}{4}\right)^{3-k} = \frac{1}{4^k} \cdot \frac{3^{3-k}}{4^{3-k}} = \frac{3^{3-k}}{4^3} = \frac{3^{3-k}}{64} $$