Explaining that $1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1)$ I have a few students that are having trouble understanding that
$$1 \cdot 3 \cdot 5 \dotsm (2n+1) = 1 \cdot 3 \cdot 5 \dotsm (2n-1)(2n+1),$$
specifically that
$$\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} = 2n+1.$$
I've tried explaining it a few different ways, but it didn't seem to go over very well.  What would be a good way to explain this?

I would suggest including more terms at the end of the products. That makes it easier to see the pattern and to see why it's just $(2n+1)$ that's left at the end.
$$
\require{cancel}
\begin{align}
\frac{1 \cdot 3 \cdot 5 \dotsm (2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-1)} &= 
\frac{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)(2n+1)}{1 \cdot 3 \cdot 5 \dotsm (2n-5)(2n-3)(2n-1)\quad\quad\quad} \\ 
&=\frac{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}(2n+1)}{\cancel{1 \cdot 3 \cdot 5} \dotsm \cancel{(2n-5)}\cancel{(2n-3)}\cancel{(2n-1)}\quad\quad\quad} \\
&=2n+1
\end{align}$$
It might also simply be a matter of spacing: students might get confused by seeing the $2n+1$ directly above the $2n-1$. I've tried to format the above to emphasize that the $2n+1$ is an extra factor that doesn't appear in the bottom.
Finally, it could also be helpful to pick a small $n$ and work out the product concretely.
