On these course notes, we are given the distribution of the posterior distribution https://people.eecs.berkeley.edu/~jordan/courses/260-spring10/lectures/lecture5.pdf
This famous result can be found in many other places, such as https://www.cs.ubc.ca/~murphyk/Papers/bayesGauss.pdf. However, I am confused about the coefficient (i.e., the $\dfrac{1}{\sqrt{2\pi}\sigma}$) term associated with the Gaussian for the posterior distribution. Here is what I am having trouble with.
We know that given $\mathcal{D} = (x_1, \ldots, x_n), x_i$ iid,
$$p(\mu|\mathcal{D}) \propto p(D|\mu)p(\mu)$$
Suppose that $$p(\mu) = \dfrac{1}{\sqrt{2\pi}\sigma_o} \exp{\dfrac{(x - \mu_o)^2}{2\sigma_o^2}}.$$ and $$p(D|\mu)= \dfrac{1}{(2\pi\sigma^2)^\frac{N}{2}} \exp(-\dfrac{1}{2\sigma^2}\sum\limits_{n = 1}^N (x_n - \mu))$$
then multiplying the expressions together, we obtain:
$$\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}} \exp{\left[-\dfrac{1}{2\sigma^2}\sum\limits_{n = 1}^N (x_n - \mu)\right]} \dfrac{1}{\sqrt{2\pi}\sigma_o} \exp{\left[\dfrac{(x - \mu_o)^2}{2\sigma_o^2}\right]}$$
While we can perform a complete the square inside of the exponential, what about the constants $\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}}$ and $\dfrac{1}{\sqrt{2\pi}\sigma_o}$?
I don't see how $\dfrac{1}{(2\pi\sigma^2)^\frac{n}{2}} *\dfrac{1}{\sqrt{2\pi}\sigma_o} = \dfrac{1}{\sqrt{2\pi}\sigma_n^2}$
where $\sigma_n^2 = (1/\sigma_o^2 + n/\sigma^2)^{-1}$ as shown in Lemma 6.
I tried playing around with the terms but I couldn't make them equal, even when $n = 1$. Have I made a mistake or is this because $p(\mu|\mathcal{D})$ is not a "true" probability distribution (i.e., doesn't integrate to 1)?
If so, how would people deal with this leading coefficient during simulation?
Addendum (see comment)
Bishop Pattern Reconigition and Machine Learning (2006) Pg. 98
