Solving a Differential Equation by Laplace Transform: $ty''(t)+(1-n)y'(t)+y(t)=0$

Question

Solve the initial value problem \begin{cases} ty''(t)+(1-n)y'(t)+y(t)=0~~~~~~~~(n=2,3,4...)\\ y(0)=0,~~~~y'(0)=0 \end{cases}

I have tried this: Laplace Transform.

$\mathcal{L}[y(t)]=Y(p)\\ \mathcal{L}[y'(t)]=pY(p)\\ \mathcal{L}[y''(t)]=p^2Y(p) $

Then: $$ty''(t)=-\frac{\mathrm d}{\mathrm dp}[p^2Y(p)]$$

Then: $$ \frac{\mathrm d Y}{Y}=\frac{p^2}{1-(n+1)p}\mathrm d p$$

But I don't know what to do then.

In my textbook, the answer is :

$$y(t)=C\sum_{k=0}^\infty\frac{(-1)^k}{k!(k+n)!}t^{k+n}$$ C is an arbitrary constant.