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Intuitive explanation for dividing in (n-1) when calculating sd?

So I've been asked today in class why you divide the sum of square error with $(n-1)$ instead of with $n$, when calculating the sd.

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?!

Tal Galili
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