jmc

Five-digit numbers will have a minimum of $1^2+2^2+3^2+4^2+5^2=55$ as the sum of their squares if the five digits are distinct and non-zero. If there is a zero, it will be forced to the left by rule #2. No digit will be greater than $7$, as $8^2=64$. Trying four digit numbers $WXYZ$, we have $w^2+x^2+y^2+z^2=50$ with $0<w<x<y<z<8$ $z=7$ will not work, since the other digits must be at least $1^2+2^2+3^2=14$, and the sum of the squares would be over $50$. $z=6$ will give $w^2+x^2+y^2=14$. $(w,x,y)=(1,2,3)$ will work, giving the number $1236$. No other number with $z=6$ will work, as $w,x,$ and $y$ would have to be greater. $z=5$ will give $w^2+x^2+y^2=25$. $y=4$ forces $x=3$ and $w=0$, which has a leading zero, and then we have 345 which is a 3-digit number. $z=4$ can only give the number $1234$, which does not satisfy the condition of the problem. Thus, the number in question is $1236$, and the product of the digits is $36$, giving $\boxed{36}$ as the answer.