Joint Density for Independent Normal Variables

$X$ is a normally distributed variable with mean, $\mu$, and known variance, $\sigma^2$:
$X_1,..., X_n \overset{iid}{\sim} N(\mu, \sigma^2)$

The joint density is:
$$\begin{align} f_{\theta}(x_1,...,x_n) &= \prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi \sigma^2}}e^{-\frac{1}{2\sigma^2}(x_i-\mu)^2} \\ &= (2 \pi \sigma^2)^\frac{n}{2} \prod_{i=1}^{n} e^{-\frac{1}{2\sigma^2}(x_i-\mu)^2} \\ &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n (x_i-\mu)^2} \\  \end{align}$$

In the exponent, use the following substitution $(x_i-\mu)^2 = x_i^2 - 2x_i\mu + \mu^2$:

$$\begin{align} f_{\theta}(x_1,...,x_n) &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n (x_i-\mu)^2} \\  &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 - 2x_i\mu + \mu^2} \\ &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 + \frac{1}{2\sigma^2} \sum_{i=1}^n 2x_i\mu -\frac{1}{2\sigma^2} \sum_{i=1}^n \mu^2} \\  &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2 + \frac{\mu}{\sigma^2} \sum_{i=1}^n x_i -\frac{n\mu^2}{2\sigma^2}} \\  &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} \sum_{i=1}^n x_i } e^{-\frac{n\mu^2}{2\sigma^2}} \\  \end{align}$$

Now set $T= \sum_{i=1}^n x_i$, then re-arrange the expression to get:

$$\begin{align} f_{\theta}(x_1,...,x_n) &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} \sum_{i=1}^n x_i } e^{-\frac{n\mu^2}{2\sigma^2}} \\ &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2} e^{\frac{\mu}{\sigma^2} T } e^{-\frac{n\mu^2}{2\sigma^2}} \\ &= (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{n\mu^2}{2\sigma^2}} e^{\frac{\mu}{\sigma^2} T } e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2}   \\ \end{align}$$

This expression can be factorized into $g_{\mu}(t) = (2 \pi \sigma^2)^\frac{n}{2} e^{-\frac{n\mu^2}{2\sigma^2}} e^{\frac{\mu}{\sigma^2} T }$ and $h(x_1,...,x_n) = e^{-\frac{1}{2\sigma^2} \sum_{i=1}^n x_i^2}$, so $T$ is a sufficient statistic for $\mu$.