# Distribution of quadratic form of normals

I am trying to figure out the distribution of
$$(n-1) \sum_{i=1}^n Z_i^2 – \left( \sum_{i=1}^n Z_i \right)^2 \qquad (*)$$
where $Z_i \sim \mathcal{N}(0,1)$, i.i.d. I know that, taking each of the terms separately,
$$\sum_{i=1}^n Z_i^2 \sim \chi^2(n)$$
and
$$\frac{1}{n}\left( \sum_{i=1}^n Z_i \right)^2 \sim \chi^2(1).$$
But I am unsure about the distribution of (*)

Here is an attempt:

Consider $Z=X-Y$ such that $X \sim \chi^2(\alpha)$ and $Y \sim \chi^2(\beta)$
with $\alpha \geq \beta$

$$\mathcal{M}_X(t) = \left(1-2 \, t\right)^{-\alpha/2}$$

$$\mathcal{M}_Y(t) = \left(1-2 \, t\right)^{-\beta/2}$$

$$\mathcal{M}_Z(t) = M_X(t)M_Y(-t) = \left(1-2 \, t\right)^{-\alpha/2}\left(1+2 \, t\right)^{-\beta/2} = (1-4t^2)^{-\beta/2}$$

$$\mathcal{M}_Z(t) = (1-2t)^{-n/2}(1+2t)^{-1/2} = (1-4t^2)^{-1/2} (1-2t)^{-(n-1)/2}$$

I am not sure if it can be reduced to a fathomable MGF.