Answer:

For \(n=1\),

\(\frac{d}{dx}(xe^{ax})=e^{ax}+axe^{ax},\)

which agrees with

\(1\cdot a^{0}e^{ax}+a^{1}xe^{ax}.\)

Assume that for some \(n=k\geq 1\),

\(\frac{d^{k}}{dx^{k}}(xe^{ax})=ka^{k-1}e^{ax}+a^{k}xe^{ax}.\)

Differentiate once more:

\(\frac{d^{k+1}}{dx^{k+1}}(xe^{ax})=ka^{k}e^{ax}+a^{k}e^{ax}+a^{k+1}xe^{ax}.\)

Hence

\(\frac{d^{k+1}}{dx^{k+1}}(xe^{ax})=(k+1)a^{k}e^{ax}+a^{k+1}xe^{ax}.\)

This is the required formula with \(n=k+1\). Since the result is true for \(n=1\) and the induction step is valid, it is true for all positive integers \(n\).