Answer
$$P_{n}(x) = x^{n} -nP_{n-1}(x) $$
\begin{align*}
P_1(x)&= x-P_0(x)=x-1\\
P_2(x)&= x^2-2P_1(x)=x^2-2x+2\\
P_3(x)&= x^3-3P_2(x)=x^3-3x^2+6x-6\\
P_4(x)&= x^4-4P_3(x)=x^4-4(x^3-3x^2+6x-6)\\
\end{align*}
Work Step by Step
Given $$ \int x^{n} e^{x} d x=P_{n}(x) e^{x}+C$$ Since we have $$ \int x^{n} e^{x} d x=x^{n} e^{x}-n \int x^{n-1} e^{x} d x$$ Then \begin{align*} \int x^{n} e^{x} d x&=x^{n} e^{x}-n \int x^{n-1} e^{x} d x\\ P_{n}(x) e^{x} &= x^{n} e^{x}-nP_{n-1}(x)e^x \\ \end{align*} Hence $$P_{n}(x) = x^{n} -nP_{n-1}(x) $$
Since $P_0(x)=1$, then
\begin{align*}
P_1(x)&= x-P_0(x)=x-1\\
P_2(x)&= x^2-2P_1(x)=x^2-2x+2\\
P_3(x)&= x^3-3P_2(x)=x^3-3x^2+6x-6\\
P_4(x)&= x^4-4P_3(x)=x^4-4(x^3-3x^2+6x-6)\\
\end{align*}
