Find a formula for the \(n^{th}\) term of the Taylor series of \(f(x)\), centered at \(c\):
\[ f(x) = \frac{1}{x}; \quad c = 1 \]
\[ f'(x) = -\frac{1}{x^2}; \quad f''(x) = \frac{2}{x^3}; \quad f'''(x) = -\frac{6}{x^4}; \quad f^{(4)}(x) = \frac{24}{x^5} \]
Now we can solve for \(f^{(n)}\):
\[ f^{(n)}(x) = (-1)^{n+1} \cdot \frac{1}{x^n} \]
\[ f^{(n)}(c) = (-1)^{n+1} \]