Answer
$f^{(n)}(x) = \frac{(2\cdot3\cdot4\cdot ...\cdot n)}{(2-x)^{n+1}}=\frac{n!}{(2-x)^{n+1}}$
Work Step by Step
$f(x) = \frac{1}{2-x}$
$f'(x) = \frac{1}{(2-x)^2}$
$f''(x) = \frac{2}{(2-x)^3}$
$f'''(x) = \frac{(2\cdot3)}{(2-x)^4}$
$f^{(4)}(x) = \frac{(2\cdot3\cdot4)}{(2-x)^5}$
etc....
$f^{(n)}(x) = \frac{(2\cdot3\cdot4\cdot ...\cdot n)}{(2-x)^{n+1}}$
