Solution: Rewriting the given differential equation as \[ \frac{\mathrm{d}y}{\mathrm{d}x} - \frac{n}{x+1} y = e^x (x+1)^n. \] The above one is a linear differential equation and this an be solved by finding the integrating factor. \begin{align*} I & = e^{\int -\frac{n}{x+1} \mathrm{d}x } = e^{-n \ln (x+1)} = (x+1)^{-n}. \end{align*} Multiplying the given differential equation with the integrating factor gives \begin{align*} & (x+1)^{-n}\frac{\mathrm{d}y}{\mathrm{d}x} - n (x+1)^{-n-1} y = e^{x} \\ \implies & \frac{\mathrm{d}}{\mathrm{d}x} \left( y (x+1)^{-n} \right) \equiv e^{x} \\ \implies & y (x+1)^{-n} = \int e^x \mathrm{d} x + c \\ \implies & y (x+1)^{-n} = e^{x} + c \\ \implies & y = (x + 1)^n e^x + c (x+1)^n. \end{align*} Now we will use the initial condition to find the value of $c$. \begin{align*} y(0) = 1 & \implies 1 = 1 + c \implies c = 0. \end{align*} Thus, the solution of the given differential equation will be \[ y = (x+1)^n e^x . \]