First of all that's a lame definition of a chi-square.

The real definition is

WHERE the dfs need not be an integer.

It's easy to prove that if you square a st normal you get a chi-square with 1 df and then via MGFs you can show that sums of independent chi-squares gives you a chi-square.

NOW, when you subtract the sample mean you do lose that 1 df.

It's not a simple proof and I couldn't find it on the web.

I'm sure it's here and I'll look again.