Let $I_n = \int_0^\pi \sin^n(x)dx$ and suppose that we have already established a recursive relation $I_n = \frac{n - 1}{n}I_{n-2}$ and we know that $\Gamma(x + 1) = x\Gamma(x), \Gamma(1/2) = \sqrt{\pi}, \Gamma(3/2) = \frac{\sqrt{\pi}}{2}$ and $I_2 = \frac{\pi}{2}$. Currently, I get a wrong end result and I don't understand where my fault lies.
Assuming that $n = 2k, k \in \mathbb{N}$, we have the identity $ I_{2k} = \frac{2k - 1}{2k}\frac{2k - 3}{2k - 2}\cdots\frac{1}{4}I_2 = \frac{\frac{2k - 1}{2}}{\frac{2k}{2}}\frac{\frac{2k - 3}{2}}{\frac{2k - 2}{2}}\cdots\frac{\frac{1}{2}}{\frac{4}{2}}\frac{\pi}{2}.$ Then, as $\Gamma\left(\frac{2k + 1}{2}\right) = \frac{2k - 1}{2}\cdot \frac{2k - 3}{2}\cdots \frac{5}{2}\cdot\frac{3}{2}\Gamma\left(\frac{1}{2}\right)$, we have the equality $\Gamma\left(\frac{n + 1}{2}\right) = 2\sqrt{\pi} \cdot \frac{2k - 1}{2}\frac{2k - 3}{2}\cdots \frac{3}{2}\cdot\frac{1}{2}$. Therefore $I_{2k} = \frac{\Gamma\left(\frac{2k + 1}{2}\right)}{\Gamma\left(\frac{2k + 2}{2}\right)2\sqrt{\pi}}\frac{\pi}{2} = \frac{\Gamma\left(\frac{2k + 1}{2}\right)\pi}{\Gamma\left(\frac{2k + 2}{2}\right)4\sqrt{\pi}} = \frac{\Gamma\left(\frac{2k + 1}{2}\right)\pi}{\Gamma\left(\frac{2k + 2}{2}\right)8\cdot\Gamma\left(\frac{3}{2}\right)}$, but a written source I am reading states that $I_{2k} = \frac{\Gamma\left(\frac{2k + 1}{2}\right)\pi}{\Gamma\left(\frac{2k + 2}{2}\right)2\cdot\Gamma\left(\frac{3}{2}\right)}$. So what am I missing?
Your idea is correct, there is just something wrong with the factors. I will try to give a detailed calculation to show that for $n\in\mathbb{N}_0$ $$ \int_0^{\pi} \sin^n(x)dx = \frac{\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}+1\right)}\sqrt{\pi}. $$
Proof. We have $$ I_{2n} = \frac{2n-1}{2n}I_{2n-2} = \dots = \left(\prod_{k=0}^{n-1}\frac{2n-(2k+1)}{2n-2k}\right)\cdot A_0 = \frac{(2n-1)(2n-3)\dots\cdot 3\cdot 1}{2n(2n-2)\dots\cdot 4\cdot 2}\cdot A_0 $$ and $$ I_{2n+1} = \frac{2n}{2n+1}I_{2n-1}=\dots= \left( \prod_{k=0}^{n-1}\frac{2n-2k}{2n-(2k-1)}\right) \cdot A_1 = \frac{2n(2n-2)\dots\cdot 4\cdot2}{(2n+1)(2n-1)\dots\cdot 5\cdot 3}\cdot A_1 $$ Let us check the expressions for $\Gamma((m+1)/2)$ and $\Gamma(m/2 + 1)$ for even and odd values of $m$ separately. The Gamma function satisfies $\Gamma(x+1)=x\Gamma(x)$. Therefore, using $\Gamma(1/2)=\sqrt{\pi}$, $$ \Gamma\left(\frac{2n+1}{2}\right) = \frac{2n-1}{2}\cdot\Gamma\left(\frac{2n-1}{2}\right) = \dots = \frac{2n-1}{2}\frac{2n-3}{2}\dots\frac{3}{2}\cdot\frac{1}{2}\cdot\sqrt{\pi} $$ where the numerators are of the form $2n-(2k+1)$ for $k\in\{0,\dots,n-1\}$. Furthermore, \begin{align} \Gamma\left(\frac{(2n+1)+1}{2}\right) &= \Gamma\left(\frac{2n+2}{2}\right) = \Gamma(n+1) = n! \\ \Gamma\left(\frac{2n}{2}+1\right) &= \Gamma(n+1)=n! \end{align} And similar to above, $$ \Gamma\left(\frac{2n+1}{2}+1\right) = \frac{2n+1}{2}\Gamma\left(\frac{2n+1}{2}\right) =\dots = \frac{2n+1}{2}\frac{2n-1}{2}\dots\frac{3}{2}\cdot \frac{1}{2}\cdot\sqrt{\pi} $$ where the numerators are of the form $2n-(2k-1)$ for $k\in\{0,\dots,n\}$.
Extending $I_{2n}$ and $I_{2n+1}$ by $2/2$ (as you did), and noting that $I_0=\pi$ and $I_1=2$, we see $$ I_{2n} = \frac{\frac{2n-1}{2}\frac{2n-3}{2}\dots\frac{3}{2}\frac{1}{2}}{\frac{2n}{2}\frac{2n-2}{2}\dots\frac{4}{3}\frac{2}{2}}\cdot \pi = \frac{\frac{2n-1}{2}\frac{2n-3}{2}\dots\frac{3}{2}\frac{1}{2}}{n(n-1)\dots 2\cdot 1}\cdot\pi =\frac{\Gamma\left(\frac{2n+1}{2}\right)}{\Gamma\left(\frac{2n}{2}+1\right)}\cdot\sqrt{\pi} $$ and, being careful here with the factors, $$ I_{2n+1} = \frac{\frac{2n}{2}\frac{2n-2}{2}\dots\frac{4}{2}\frac{2}{2}}{\frac{2n+1}{2}\frac{2n-1}{2}\dots\frac{5}{3}\frac{3}{2}} \cdot 2 = \frac{n(n-1)\dots 2\cdot 1}{2\Gamma\left(\frac{2n+1}{2}+1\right)} \cdot\sqrt{\pi}\cdot 2 = \frac{\Gamma\left(\frac{(2n+1)+1}{2}\right)}{\Gamma\left(\frac{2n+1}{2}+1\right)}\cdot \sqrt{\pi}\,. $$ This proves the claim.