Sufficient statistic for $N(\mu,1)$ Let $X_1,\ldots,X_n$ be a random sample from $N(\mu,1)$, where $\mu$ is an unknown parameter. Show that $(\overline{x}/{S^2}, S^2)$ is a sufficient statistic for $\mu$, where $S^2$ is the sample variance.
My Approach
$$
\sum_{i=1}^n (x_i-\mu)^2 = n(\bar x-\mu)^2 + \sum_{i=1}^n (x_i-\bar x)^2.
$$
By Neyman Fisher Factorisation,
\begin{align}
f_{X_1,\ldots,X_n}(x_1,\ldots,x_n) \propto {} & \prod_{i=1}^n \frac 1 {(\sqrt2\pi)^n} \exp\left( \frac{-1} 2 \left( {x_i-\mu} \right)^2 \right) = \frac 1 {(\sqrt2\pi)^n} \exp\left( \frac{-1}{2} \sum_{i=1}^n (x_i-\mu)^2  \right) \\[10pt]
= {} & (\sqrt2\pi)^{-n} \exp\left( \frac{-1}{2} \left( n(\bar x - \mu)^2 + \sum_{i=1}^n (x_i - \bar x)^2 \right) \right) \\[10pt]
= {} & (\sqrt2\pi)^{-n} \exp\left( - \frac n {2} ((\bar x - \mu)^2 + s^2) \right).
\end{align}
How do I proceed from here?
 A: Since the variance in this problem is a known constant, it is actually the case that $\bar{x}$ is sufficient for $\mu$, and since $\bar{x}$ can be obtained from the stated statistic (as the product of the two elements), that statistic is also sufficient for $\mu$.  To see that $\bar{x}$ is sufficient for $\mu$ we can apply the Fisher-Neyman factorisation theorem just as you have attempted to do.  In this problem your likelihood function is given by:$^\dagger$
$$\begin{equation} \begin{aligned}
L_\mathbf{x}(\mu) 
&= (2 \pi)^{-n/2} \cdot \exp \Big( - \frac{1}{2} (x_i-\mu)^2 \Big) \\[6pt]
&= (2 \pi)^{-n/2} \cdot \exp \Big( - \frac{n}{2} \cdot (\bar{x}-\mu)^2 - \frac{n-1}{2} \cdot s^2 \Big) \\[6pt]
&= \underbrace{(2 \pi)^{-n/2} \cdot \exp \Big( - \frac{n-1}{2} \cdot s^2 \Big)}_{h(\mathbf{x})} \cdot \underbrace{\exp \Big( - \frac{n}{2} \cdot (\bar{x}-\mu)^2 \Big)}_{g_\mu(\bar{x},n)}. \\[6pt]
\end{aligned} \end{equation}$$
It follows that $\bar{x}$ is sufficient for $\mu$.  With some more work it is possible to show that it is minimal sufficient, but that is unnecessary for the result you want in this case.

$^\dagger$ I have used the standard notation where the sample variance uses Bessel's correction) so my result is slightly different to yours.
