$\displaystyle x^n-c^n=(x^\frac{n}{2}-c\frac{n}{2})$

I like to use the calculator to double check myself, but for some reason I cannot produce the correct answer for the problem below.

$\displaystyle x^8-\frac{1}{16}=(x^4+\frac{1}{4})(x^4+\frac{1}{4}) $

$\displaystyle =(x^2-\frac{1}{2})(x^2+\frac{1}{2})(x^4+\frac{1}{4})$

The first step I attempt to get $\displaystyle (x^4+\frac{1}{4})(x^4+\frac{1}{4}) $ by putting this into the calculator$\displaystyle (\frac{1}{16})^{(\frac{1}{4})}$ which results in .5 or 1/2. How do I keep on missing the first step to the solution which is the 1/4?