If T follows a t-distribution them prove that U=T^2 follows an F-distribution 
Is this correct is there a more elegant way to do this 
 A: The argument is correct but some of the notation is not standard and I would be somewhat more verbal.
In particular, I've never see the notation $N^2(0,1)$ before.
You wrote:

If $T$ has a t-distribution then $$ T = \frac Z {\sqrt{W/\nu}} $$ where $Z\sim N(0,1)$ and $W\sim\chi^2(\nu).$

It's not clear whether it says $\sqrt{W/\nu}$ or $\sqrt W/n$, so that's a small point to address, which could be consequential as far as the reader's understanding is concerned. A more substantial issue is that I would explicitly mention that the numerator and the denominator are independent. And there's something about which I have a pet peeve: The word "where" should be used for defining notation, but should not be used in effect as a logical quantifier, in this case meaning "for some" random variables $Z\sim N(0,1)$ and $W\sim\chi^2(\nu).$" That could be the place where you mention independence explicitly, thus phrasing it as "$\ldots\,$for some independent random variables $Z\sim N(0,1)$ and $W\sim\chi^2(\nu).$"
Then you could say "$T^2= \dfrac{Z^2/1}{W/\nu} = \dfrac{\chi^2(1)/1}{\chi^2(\nu)/\nu},$ where the numerator and the denominator are independent. Therefore $T^2\sim F(1,\nu).$" And "$T^2=F(1,\nu)$" should say instead "$T^2\sim F(1,\nu).$"
