Base case: For \(n = 1\),

\(\frac{d}{dx} \left( x^2 e^x \right) = x^2 e^x + 2x e^x = (x^2 + 2x) e^x.\)

So true when \(n = 1\).

Inductive step: Assume that

\(\frac{d^k}{dx^k} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x\)

for some value of \(k\).

Then,

\(\frac{d^{k+1}}{dx^{k+1}} \left( x^2 e^x \right) = \left( x^2 + 2kx + k(k-1) \right) e^x + e^x (2x + 2k).\)

Simplifying gives:

\(\left( x^2 + 2(k+1)x + k(k+1) \right) e^x.\)

So true when \(n = k+1\). By induction, true for all positive integers \(n\).