How to calculate this integral using Rodrigues' formula?

I'm trying to get practice using Rodrigues' formula for Legendre Polynomials, but it's being quite confusing to manipulate that $n$-th derivative. Basically, I'm trying to calculate:

$$\int_{-1}^1 x^n P_n(x) dx$$

using Rodrigues' formula. Substituing $P_n$ we have

$$\int_{-1}^1 x^n P_n(x)dx = \dfrac{1}{2^n n!} \int_{-1}^1 x^n \dfrac{d^n}{dx^n}(x^2-1)^ndx$$

Now integrating by parts I have the following:

$$\int_{-1}^1 x^nP_n(x)dx=\dfrac{1}{2^n n!}\left(x^n \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^n \bigg|_{-1}^1 - n\int_{-1}^1 x^{n-1} \dfrac{d^{n-1}}{dx^{n-1}}(x^2-1)^ndx\right)$$

Now I don't know how to evaluate that boundary term, also I would need to do integration by parts again, but then what? I imagine I would keep doing it until the derivative disappears. In the process I would get a lot of boundary terms and then in the end

$$\int_{-1}^1x^nP_n(x)dx = \dfrac{1}{2^n n!}\left([\text{Boundary terms}] + (-1)^n n!\int_{-1}^1(x^2 -1)^n dx\right)$$

Where the $(-1)^n$ comes because at each step the new integral is multiplied by the sign of the last one and the $n!$ appears because the $n$ on the first integral multiplies $(n-1)$ on the second and so forth.

But still, this is something I see intuitively, how can I really calculate that? Also, how to deal with all those boundary terms? How to make this computation in a good way?

let's look at those boundary terms, keeping the Product rule for higher derivatives in mind (let $n<k$ and $m=k-n$) \begin{align*} \frac{d\,^n}{d x^n}\left(\left(x^2-1\right)^k\right) &= \frac{d\,^n}{d x^n}\left((x-1)^k(x+1)^k\right) \\ &= \sum_{i=0}^n \binom{n}{i} \frac{d\,^{i}}{d x^{i}}(x-1)^k\frac{d\,^{n-i}}{d x^{n-i}}(x+1)^{k} \\ &= \sum_{i=0}^n \binom{n}{i} \frac{k!}{(k-i)!}(x-1)^{k-i}\frac{k!}{(k-n+i)!}(x+1)^{k-n+i} \\ &= \sum_{i=0}^n \binom{n}{i} \frac{(k!)^2}{(k-i)!(k-n+i)!}(x-1)^{k-i}(x+1)^{k-n+i}\\ &= \sum_{i=0}^n \binom{n}{i} \frac{(k!)^2}{(k-i)!(k-n+i)!}(x-1)^{m+n-i}(x+1)^{m+i}\\ &= (x^2-1)^m\sum_{i=0}^n \binom{n}{i} \frac{(k!)^2}{(k-i)!(m+i)!}(x-1)^{i}(x+1)^{k-n+i} \end{align*}
As we see for all $n<k$ the derivative are zero for $x=\pm 1$