The quoted formula is not quite right. Let's derive the correct one.
Since the population mean (or any other constant) may be subtracted from every value in a population $S$ without changing the variance of the population or of any sample thereof, we might as well assume the population mean is zero. Letting the values in the population be $\{x_i\, \vert\, i\in S\}$, this implies
$$0 = \sum_{i\in S} x_i.$$
Squaring both sides maintains the equality, giving
$$0 = \sum_{i,j\in S}x_ix_j = \sum_{i\in S}x_i^2 + \sum_{i \ne j \in S} x_ix_j,$$
whence
$$\sum_{i\ne j \in S} x_ix_j = -\sum_{i\in S} x_i^2.$$
This key result will be employed later.
Let $S$ have $N$ elements. Because its mean is zero, its variance is the average squared value:
$$s^2 = \frac{1}{N}\sum_{i\in S}x_i^2.$$
(Please note that there can be no dispute about the denominator of $N$; in particular, it definitely is not $N-1$: this is a population variance, not an estimator.)
To find the variance of the sample distribution of the mean, consider all possible $n$-element samples. Each corresponds to an $n$-subset $A\subset S$ and has mean
$$\frac{1}{n}\sum_{i\in A} x_i.$$
Since the mean of all the sample means equals the mean of $S$, which is zero, the variance of these $\binom{N}{n}$ sample means is the average of their squares:
$$s_n^2 = \frac{1}{\binom{N}{n}} \sum_{A\subset S}\left(\frac{1}{n}\sum_{i\in A}x_i\right)^2 = \frac{1}{n^2\binom{N}{n}} \sum_{A\subset S}\sum_{i,j\in A}x_ix_j \\= \frac{1}{n^2\binom{N}{n}} \sum_{A\subset S}\left(\sum_{i\in A}x_i^2 + \sum_{i\ne j\in A}x_ix_j\right) .$$
(Once again, $\binom{N}{n}$, not $\binom{N}{n}-1$, is the correct denominator: this is the variance of a collection of $\binom{N}{n}$ numbers, not an estimator of anything.)
Fix, for a moment, any particular index $i$. The value $x_i$ will appear in $\binom{N-1}{n-1}$ samples, because each such sample supplements $x_i$ with $n-1$ more elements of $S$ out of the $N-1$ remaining elements (sampling is without replacement, remember). Its contribution to the right hand side therefore equals $\binom{N-1}{n-1}x_i^2$.
Also fixing an index $j\ne i$, similar reasoning shows the product $x_ix_j$ appears in $\binom{N-2}{n-2}$ samples, thereby contributing $\binom{N-1}{n-1}x_ix_j$ to the right hand side. Therefore, upon summing over all such $i$ and $j$ in $S$,
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1}\sum_{i\in S}x_i^2 + \binom{N-2}{n-2}\sum_{i\ne j\in S}x_ix_j\right).$$
Plug the first result into that last sum:
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1}\sum_{i\in S}x_i^2 + \binom{N-2}{n-2}\left(-\sum_{i\in S}x_i^2\right)\right).$$
It is now straightforward to relate this to the variance of $S$, because $\sum_{i\in S}x_i^2 = Ns^2$:
$$s_n^2 = \frac{1}{n^2\binom{N}{n}} \left(\binom{N-1}{n-1} - \binom{N-2}{n-2}\right)\left(Ns^2\right) = \frac{s^2}{n}\left(1 - \frac{n-1}{N-1}\right).$$
Thus the sampling variance for sampling with replacment, $\frac{s^2}{n}$, is multiplied by $1 - \frac{n-1}{N-1}$ to obtain the sampling variance for sampling without replacement, $s_n^2$. Accordingly, the multiplicative adjustment for the sampling standard deviation is its square root, $\sqrt{1- \frac{n-1}{N-1}}$. This differs from the quoted formula, which uses $\sqrt{1 - \frac{n}{N}}$.
Two simple checks can give us some comfort concerning the correctness of this result. First, the sample variance of means of samples of size $n=1$, $s_1^2$, obviously equals the population variance $s^2$. The correct formula states
$$s_1^2 = \frac{s^2}{1}\left(1 - \frac{1-1}{N-1}\right) = s^2,$$
as it should. Unfortunately, the quoted formula asserts that $s_1^2 = s^2(\frac{1}{1} - \frac{1}{N})$ which obviously cannot be right. Second, the sample variance of the means of samples of size $n=N$ is zero, because there is no variation, and indeed both formulas give $0$ in this case.
