One way to get at this is to assume \(\displaystyle \sigma^2 = 1\), prove it for the case n = 2, then induct by showing that

\(\displaystyle (n - 1) S^2_n = (n - 2) S^2_{n - 1} + \left(\frac{n - 1}{n}\right)(X_n - \bar{X}_{n - 1})^2\),

which will get you where you need since you will have the sum of two independent chi-squares. Some facts that you will use are that the sum of independent chi-squares is chi-square, that the square of an \(\displaystyle N(0, 1)\) is \(\displaystyle \chi^2_1\), and that \(\displaystyle \bar{X}_n \perp S^2_n\). The definitions are:

\(\displaystyle S^2 _n = \frac{\sum_{i = 1} ^ n (X_i - \bar{X}_{n})^2}{n - 1}

\)

\(\displaystyle \bar{X}_n = \sum_{i = 1} ^ n \frac{X_i}{n}\)

Once you have this, it's just a matter of undoing the assumption that \(\displaystyle \sigma^2 = 1\), which isn't really a big deal. To be honest, if this is just for kicks, it's actually more of a pain in the head than it appears, particularly in proving the identity above.