Prove, by induction, that the nth derivative of f(x) = x^2 e^x is f^(n)(x) = (x^2 + 2nx + n(n-1))e^x

Prove, by induction, that the nth derivative of f(x) = x^2 e^x is f^(n)(x) = (x^2 + 2nx + n(n-1))e^x

Any help would be much appreciated
Thanks

Base Case:
This is true for n = 1, because f '(x) = x^2 e^x + 2x e^x = (x^2 + 2 * 1x + 0) e^x.

Inductive Step:
Assume that f^(k)(x) = (x^2 + 2kx + k(k-1)) e^x.

So, f^(k+1)(x)
= (d/dx) [(x^2 + 2kx + k(k-1)) e^x], by the inductive hypothesis
= (x^2 + 2kx + k(k-1)) e^x + (2x + 2k) e^x, by the product rule
= [(x^2 + 2kx + k(k-1)) + (2x + 2k)] e^x, by factoring
= [x^2 + (2k+2)x + (k^2+k)] e^x
= [x^2 + 2(k+1)x + (k+1)k] e^x, which establishes the claim for n = k+1.

Hence, the nth derivative of f(x) = x^2 e^x is f^(n)(x) = (x^2 + 2nx + n(n-1))e^x for all positive integers n.

I hope this helps!