use the formula $P_n(x) = \dfrac{1}{2^nn!}\dfrac{d^n}{dx^n}((x^2-1)^n)$ to show that $P_{2n}(0) = \dfrac{(-1)^n(2n)!}{4^n(n!)^2}$ and odd terms are 0.
I first subbed in 2n to the formula and got
$P_{2n}(x) = \dfrac{1}{4^n(2n)!} \dfrac{d^{2n}}{dx^{2n}}((x^2-1)^{2n})$ but I am not sure how to deal with differentiating that term $2n$ times. I have tried to use the binomial theorem but to no avail.
By the binomial theorem, $(x^2-1)^{2n} = \displaystyle\sum_{k = 0}^{2n}\dbinom{2n}{k}(x^2)^k(-1)^{2n-k} = \displaystyle\sum_{k = 0}^{2n}(-1)^{2n-k}\dbinom{2n}{k}x^{2k}$.
Now, consider $3$ cases:
If $2k < 2n$, then the $2n$-th derivative of $x^{2k}$ will be $0$.
If $2k > 2n$, then the $2n$-th derivative of $x^{2k}$ will be $\text{const} \cdot x^{2k-2n}$, which is zero when $x = 0$.
If $2k = 2n$, then the $2n$-th derivative of $x^{2k}$ will be the constant $(2n)!$.
Based on this, what is the $2n$-th derivative of $\displaystyle\sum_{k = 0}^{2n}(-1)^{2n-k}\dbinom{2n}{k}x^{2k}$ at $x = 0$?