Leibniz's formula

Use Leibniz's formula to establish that:
$\displaystyle
Z_n = \frac{{d^n }}{{dx^n }}e^{ - \frac{{x^2 }}{2}}
$
is a solution of the differential equation $\displaystyle
\frac{{d^2 Z_n }}{{dx^2 }} + x\frac{{dZ_n }}{{dx}} + (n + 1)Z_n = 0
$

I know that Leibniz's formula gives the nth derivative of a product of functions, but in this case $\displaystyle
e^{ - \frac{{x^2 }}{2}}
$ isn't a product.

So far I've got:
$\displaystyle
\begin{array}{l}
y = e^{ - \frac{{x^2 }}{2}} \\
\frac{{dy}}{{dx}} = - xe^{ - \frac{{x^2 }}{2}} \\
\frac{{d^2 y}}{{dx^2 }} = 3xe^{ - \frac{{x^2 }}{2}} - x^3 e^{ - \frac{{x^2 }}{2}} \\
\end{array}
$
but I'm at a loss as to how to generalise to n.

Help would be hugely appreciated.

first of all your second derivative is wrong. The correct expression is:

$\displaystyle \frac{{d^2 }}
{{dx^2 }}e^{ - \frac{{x^2 }}
{2}} = \left( {x^2 - 1} \right)e^{ - \frac{{x^2 }}
{2}}$

so after substituting into the differential equation we get:

$\displaystyle \frac{{d^n }}
{{dx^n }}\left( {\left( {x^2 - 1} \right)e^{ - \frac{{x^2 }}
{2}} } \right) - x\frac{{d^n }}
{{dx^n }}xe^{ - \frac{{x^2 }}
{2}} + (n + 1)\frac{{d^n }}
{{dx^n }}e^{ - \frac{{x^2 }}
{2}} = 0$

HINT: $\displaystyle {\left( {x^2 - 1} \right)}$ has only 2 derivatives before it vanishes. and x has only one....