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 (*)

Answer

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.

Attribution
Source : Link , Question Author : Zailei Chen , Answer Author : rightskewed

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