If $\theta$ is a constant such that $0 < \theta < \pi$ and $x + \dfrac{1}{x} = 2\cos{\theta}$, then for each positive integer $n$, $x^n + \dfrac{1}{x^n}$ equals
$\textbf{(A)}\ 2\cos\theta\qquad \textbf{(B)}\ 2^n\cos\theta\qquad \textbf{(C)}\ 2\cos^n\theta\qquad \textbf{(D)}\ 2\cos n\theta\qquad \textbf{(E)}\ 2^n\cos^n\theta$
Multiply both sides by $x$ and rearrange to $x^2-2x\cos(\theta)+1=0$. Using the quadratic equation, we can solve for $x$. After some simplifying:
\[x=\cos(\theta) + \sqrt{\cos^2(\theta)-1}\]
\[x=\cos(\theta) + \sqrt{(-1)(\sin^2(\theta))}\]
\[x=\cos(\theta) + i\sin(\theta)\]
Substituting this expression in to the desired $x^n + \dfrac{1}{x^n}$ gives:
\[(\cos(\theta) + i\sin(\theta))^n + (\cos(\theta) + i\sin(\theta))^{-n}\]
Using DeMoivre's Theorem:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(-n\theta) + i\sin(-n\theta)\]
Because $\cos$ is even and $\sin$ is odd:
\[=\cos(n\theta) + i\sin(n\theta) + \cos(n\theta) - i\sin(n\theta)\]
$=\boxed{\textbf{2\cos(n\theta)}},$ (Error compiling LaTeX. Unknown error_msg)
which gives the answer $\boxed{\textbf{D}}.$