I have already proved other properties of the Legendre polynomials, like:
$$P_n(-x) = (-1)^n \, P_n(x)$$ $$P_{2n+1}(0) = 0$$ $$P_n(\pm1)= (\pm1)^n$$
But I can't get this one:
$$P_{2n}(0) = \frac{(-1)^n(2n-1)!}{2^{2n-1}n((n-1)!)^2}$$
I'd be glad if someone can help me with this.
One method is to note that the Legendre polynomials can be written as the series $$ P_{n}(x) = \frac{1}{2^n} \, \sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor} (-1)^k \, \binom{n}{k} \, \binom{2n - 2k}{n} \, x^{n - 2k}. $$ Note that, when $n \to 2 n + 1$, the series is $$ P_{2n+1}(x) = \frac{1}{2^{2n+1}} \, \sum_{k=0}^{n} (-1)^k \, \binom{2n+1}{k} \, \binom{4n-2k+2}{n} \, x^{2n - 2k +1}. $$ This series gives $P_{2n+1}(0) = 0$ as a result. When $x \to -x$ it is determined that $P_{n}(-x) = (-1)^n \, P_{n}(x)$.
Now, when $n \to 2n$ the series yields $$ P_{2 n}(x) = \frac{1}{4^n} \, \sum_{k=0}^{n} (-1)^k \, \binom{2n}{k} \, \binom{4n - 2k}{n} \, x^{2(n-k)} $$ or $$ P_{2 n}(x) = \frac{1}{4^n} \, \left( (-1)^n \, \binom{2n}{n} \, \binom{2n}{2n} + \sum_{k=0}^{n-1} (-1)^k \, \binom{2n}{k} \, \binom{4n - 2k}{n} \, x^{2(n-k)} \right) $$ and, for $x = 0$, gives $$ P_{2 n}(0) = \frac{(-1)^n}{4^n} \, \binom{2n}{n} = \frac{(-1)^n}{2^{2n-1}} \, \frac{(2n-1)!}{n! \, (n-1)!}. $$ This is the result desired for the proposed question.
Also note that in order to show $P_{n}(\pm 1) = (\pm 1)^n$ that $$ \sum_{k=0}^{\lfloor{\frac{n}{2}}\rfloor} (-1)^k \, \binom{n}{k} \, \binom{2n -2k}{n} = 2^n $$ first must be obtained.