Let \(f(x) = f'(x)\) and let \( g(x) = f(x)/e^x \), then:
\[g'(x)= \frac{e^x f'(x) - f(x) e^x }{e^{2x}} \]
Simplify:
\[g'(x)= \frac{ f'(x) - f(x) }{e^x} = \frac{ f(x) - f(x) }{e^x} = 0 \]
If \( g'(x) =0 \), then \(g(x) = f(x)/e^x \) is a constant, meaning \(f(x)=Ae^x\). If \(f(x)\) would have any solutions other than \( Ae^x \), then \( g'(x)\) would not be \(0 \).