Incremental Calculation of the Sample Covariance The formula to calculate the sample covariance given $n$ vector samples $x_{i}$ for $i = 1, \ldots, n$ is as follows:
\begin{align*}
S &= \frac{1}{n-1}\sum\limits_{i=1}^{n}(x_{i} - m)(x_{i} - m)^\intercal
\end{align*}
where $m$ is defined as the sample mean:
\begin{align*}
m &= \frac{1}{n}\sum\limits_{i=1}^{n}x_{i}
\end{align*}
I have to prove that the formula for $S$ can be rewritten as:
\begin{align*}
S &= \frac{\left(\sum\limits_{i=1}^{n} x_{i} x_{i}^\intercal\right) - n\ mm^\intercal}{n - 1}
\end{align*}
I have started my proof off as follows:
\begin{align*}
S &= \frac{1}{n-1}\sum\limits_{i=1}^{n}(x_{i} - m)(x_{i} - m)^\intercal \\
&= \frac{\sum\limits_{i=1}^{n}(x_{i} - m)(x_{i} - m)^\intercal}{n - 1} \\
&= \frac{\sum\limits_{i=1}^{n}(x_{i}x_{i}^\intercal - x_{i}m^\intercal -mx_{i}^\intercal + mm^\intercal)}{n - 1} \\
\end{align*}
This is where I do not know how to proceed further. To me it seems that the only way to proceed would be to show that $x_{i}m^\intercal = mx_{i}^\intercal = mm^\intercal$, but I do not think this equality makes sense..

You have
$$
\sum\limits_{i=1}^n \big(x_i x_i^\intercal - x_i m^\intercal -mx_i^\intercal + mm^\intercal\big)
$$
with everything inside the parentheses INSIDE the summation sign.  In particular, you need to use the facts that
$$
\sum_{i=1}^n mm^\top = n m m^\top
$$
and
$$
\sum_{i=1}^n m x_i^\top = m\sum_{i=1}^n x_i^\top = m\big(nm^\top\big)
$$
and similarly for the term that is the transpose of that.
