IA: n=1
dann ist f'(x)=\(2x\cdot e^{x}+x^{2}e^{x}\)
IV: \(f^{(n)}(x)=x^{2}+2nx+(n^2-n)\cdot e^{x}\)
IS: n->n+1
\(f^{n+1}(x)=\frac{d}{dx}(x^{2}+2nx+(n^{2}-n))\cdot e^{x}=(2x+2n)e^{x}+(x^{2}+2nx+(n^{2}-n))e^{x}=(x^{2}+2x+2nx+(n^{2}-n)+2n)e^{x}=(x^{2}+2x(n+1)+((n+1)^{2}-(n+1)))e^{x}\)