Posterior of one observation transform into posterior of several observations Suppose $\mu$ has prior distribution $\mathcal{N}(M, A)$ and $x |\mu \sim \mathcal{N}(\mu, 1)$
After one observation, the posterior is $$\mu|x \sim \mathcal{N}(M + B(x-M), B), \tag{1}$$ where $B \overset{\text{(def)}}{=} \dfrac{A}{A+1}$.
After $n$ independent observations $\mathbf{x} \overset{\text{(def)}}{=}x_1, \dotsc, x_n | \mu\overset{\perp \!\!\perp}{\sim} \mathcal{N}(\mu, 1)$, the posterior is $$\mu|\mathbf{x} \sim \mathcal{N}\left(M + B_n(\bar{x} - M), \frac{1}{n}B_n \right), \tag{2}$$ where $B_n \overset{\text{(def)}}{=} \dfrac{nA}{nA + 1}$.
Is there the general method to calculate the posterior after several observations, given the prior and the posterior after one observation? Is there a path that can take me from <(1) , prior> to (2) in different situations?
It should be just manipulating Bayes' rule...someone help me out here. Thanks

Ah! Beautiful question. First, let's look at the mean value of the posterior distribution for one sample vs $n$ samples.
\begin{align}
B &= \frac{A}{A + 1} \\
B_n &= \frac{nA}{nA + 1} = \frac{A}{A + \frac{1}{n}}
\end{align}
where in the second equation I have divided the numerator and denominator by $n$. Note that this does not change the expression of $B_n$. Let's begin with one sample case and look at the mean value of the posterior distribution
\begin{align}
M + B(\bar{x} - M) = \frac{M}{A+1} + \frac{A \bar{x}}{A + 1},
\end{align}
where $A$ is the variance of the prior distribution. The above expression is a weighted average of the prior mean $M$ and the true sample mean $\bar{x}$, where the weights $1/(A + 1)$ and $A/(A+1)$ determine how much I should trust my sample mean vs prior mean.
Now let us consider the case where we collect large number samples. As $n \rightarrow \infty$, $1/n \rightarrow 0$. Therefore, $B_n \rightarrow 1$. Hence, the mean value of the posterior distribution converges to the sample mean $\bar{x}$. This is given as follows:
\begin{align}
M + B_n(\bar{x} - M) &\approx M + (\bar{x} - M) \\
&= \bar{x}
\end{align}
So, as I get more samples, I can estimate the mean by taking an average of the samples $\bar{x}$ instead of relying on the prior distribution mean $M$. In some sense, I become less dependent on the prior distribution sample mean as I get more samples. In the same way, you can also look at what happens to the variance of the posterior distribution as you get more samples. I will leave that as an exercise for you to solve.
