Answer
$$f^{(n)}(x)=(x+n)e^{x} $$
Work Step by Step
Given $$f(x)= xe^{x},\ \ f^{(n)}(x)=(x+n)e^{x} $$
At $n=1$
$$f'(x)= xe^x+e^x=(x+1)e^{x}$$
Let at $n=k$,
$$f^{(k)}(x)=(x+k)e^{x} $$
Differentiate with respect to $x$
\begin{align*}
f^{(k+1)}(x)&=(x+k)e^{x}+e^{x}\\
&=(x+k+1)e^{x}
\end{align*}
Thus is true for $k=n+1$ and
$$f^{(n)}(x)=(x+n)e^{x} $$