I was asked today in class why you divide the sum of square error by $n-1$ instead of with $n$, when calculating the standard deviation.

I said I am not going to answer it in class (since I didn't wanna go into unbiased estimators), but later I wondered - is there an intuitive explanation for this?!

The standard deviation calculated with a divisor of $n-1$ is a standard deviation calculated from the sample as an estimate of the standard deviation of the population from which the sample was drawn. Because the observed values fall, on average, closer to the sample mean than to the population mean, the standard deviation which is calculated using deviations from the sample mean underestimates the desired standard deviation of the population. Using $n-1$ instead of $n$ as the divisor corrects for that by making the result a little bit bigger.
Note that the correction has a larger proportional effect when $n$ is small than when it is large, which is what we want because when n is larger the sample mean is likely to be a good estimator of the population mean.
When the sample is the whole population we use the standard deviation with $n$ as the divisor because the sample mean is population mean.
(I note parenthetically that nothing that starts with "second moment recentered around a known, definite mean" is going to fulfil the questioner's request for an intuitive explanation.)
