I use notation from the Wikipedia page on the Orthogonal group. I seem to have found a contradiction which doesn't make sense and would like to know where I've gone wrong.
Suppose $-1$ is a square in $\mathbb{F}_q$ for some odd prime power $q$. According to the Wikipedia page:
$$|\mathrm{O}(2n,q)|=2(q^n-1)\prod_{i=1}^{n-1}(q^{2n}-q^{2i})$$ $$|\mathrm{O}(2n+1,q)|=2q^n\prod_{i=0}^{n-1}(q^{2n}-q^{2i})$$
We can embed $\mathrm{O}(2n-1,q)$ in $\mathrm{O}(2n,q)$ by $\phi\mapsto\left(\begin{array}{ll} 1 & 0 \\ 0 & \phi\end{array}\right)$ so $|\mathrm{O}(2n-1,q)|$ divides $|\mathrm{O}(2n,q)|$. However for $n>3$ $$\begin{array}{ll} \frac{|\mathrm{O}(2n,q)|}{|\mathrm{O}(2n-1,q)|} & =\frac{2(q^n-1)\prod_{i=1}^{n-1}(q^{2n}-q^{2i})}{2q^{n-1}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})} \\ & =\frac{(q^n-1)\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}{q^{n-3}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})} \\ & =\frac{(q^n-1)}{q^{n-3}} \\ \end{array}$$
which makes no sense as $q$ does not divide $q^n-1$.
Presumably I've made an obvious mistake that I can't see. What mistake have I made?
When you change the product in the numerator of $$\frac{2(q^n-1)\prod_{i=1}^{n-1}(q^{2n}-q^{2i})}{2q^{n-1}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}$$ to be a product of factors of the form $q^{2n-2}-q^{2i}$, you are factoring out $q^2$ from every single factor, not just one of them. So you get $$\frac{2(q^n-1)q^{2n-2}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}{2q^{n-1}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}=\frac{(q^n-1)q^{n-1}\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}{\prod_{i=0}^{n-2}(q^{2n-2}-q^{2i})}=(q^n-1)q^{n-1}.$$