Method of moment generating function normal distribution

Suppose that $\displaystyle Y$ is normally distributed with mean 0, and unknown variance $\displaystyle \sigma^2$. Using the method of moment generating functions, show that $\displaystyle \frac{Y^2}{\sigma^2}$ has a $\displaystyle \chi^2$ distribution with 1 df.

I know that the mgf of a normal distribution is $\displaystyle \exp\left(\mu t +\frac{t^2 \cdot \sigma^2}{2} \right) $, so if I was to replace the variable with what was given in the question then I would get:

$\displaystyle \exp\left(0 t +\frac{(Y^2/\sigma^2)^2 \cdot \sigma^2}{2} \right) = \exp\left(\frac{Y^4}{2\sigma^2} \right)$ which looks nothing like the mgf of a $\displaystyle \chi^2$ distribution.

the other way I was thinking of doing solving this was using the definition, so I would get:

$\displaystyle m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y-\mu)^2\right] \cdot \exp\left(\frac{ty^2}{\sigma^2}\right) \ dy$

$\displaystyle m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{1}{2\sigma^2}\right)(y)^2 \cdot \left(\frac{ty^2}{\sigma^2}\right)\right] \ dy$

$\displaystyle m_{Y^2/\sigma^2} (t) = E[e^{t \cdot Y^2/\sigma^2}] = \int^{+\infty}_{-\infty} \frac{1}{\sigma\sqrt{2\pi}}\cdot \exp \left[ -\left(\frac{ty^4}{2\sigma^4}\right)\right] \ dy$

at which point I don't know how to proceed