There is no requrment to know the values!

Note that,
$ABCD | DCBA$ So,
$ABCD <= DCBA$
And
$ABCD < 1111$ [Because otherwise $ABCD \times 9$ will be a 5 digit number]
So,
B, C or D will be 0 .
So,
$A \times B \times C \times D = \boxed{0}$ .

$\begin{aligned} \overline{ABCD} \times 9 & = \overline{DCBA} \\ \Rightarrow \overline{ABCD} \times (10-1) & = \overline{DCBA} \\ \overline{ABCD0} - \overline{ABCD} & = \overline{DCBA} \end{aligned}$

$\begin{array}{c}\space & A & B & C & D & 0 \\ - & & A & B & C & D \\ \hline & & D & C & B & A \end{array}$

From the above, for $\overline{AB} - A = D \text{ or } \overline{AB} - A -1 = D$ $\Rightarrow A$ must be $1$ . And for $10 - D = A = 1 \quad \Rightarrow D = 9$ .

$\begin{array}{c}\space & 1 & B & C & 9 & 0 \\ - & & 1 & B & C & 9 \\ \hline & & 9 & C & B & 1 \end{array}$

For $\begin{cases} \overline{1B} - 1 = 9 & \Rightarrow B = 0 & \Rightarrow 8 - C = 0 & \Rightarrow C = 8 \\ & \Rightarrow C-B=C & \Rightarrow 8-0= 0 & \Rightarrow \text{Acceptable} \\ \overline{1B} - 1 -1 = 9 & \Rightarrow B = 1 & \Rightarrow 8 - C = 1 & \Rightarrow C = 7 \\ & \Rightarrow C-B=C & \Rightarrow 7-1= 6 & \Rightarrow B = \text{1 and 6} \end{cases}$

Therefore, $\overline{ABCD} = 1089$ and $1089 \times 9 = 9801$ $\Rightarrow A\times B \times C \times D = \boxed{0}$

From multiplying $A \times 9$ (and adding the carry from the previous multiplications) we get a single digit $D$ . This means that $A \in \{0, 1\}$ (otherwise we would have a 2 digit number).

Since we are told that $A \neq 0$ then $A = 1$ , so $A \times 9 = 9$ and there is no carry from the multiplication of $BCD \times 9$ . This implies that $D = 9$ .

Now, $BCD \times 9 = 9 \times D + BC0 \times 9 = 81 + BC \times 90$ But the only way for this to be a 3 digit number (namely have no carry) is by making either $B,C = 0$

Finally $A \times B \times C \times D = 0$

1. A+B+C+D is a multiple of 9 (since DCBA is multiple of 9)
2. A can only be 1 (anything higher will result in D being 2 digits). So, D has to be 9.
3. B times 9 is also a single digit answer (so, B=0 or 1), any carry and A times 9 plus the carry will no longer be D.
4. A+D=10. B+C has to be 8 or 17 (assuming B or C can be 9 and 8) and B=0 or 1. So, C=8 or 7.
5. so the possible numbers are 1089 or 1179. 1179 times 9 > 10000. So, 1089 is the number and 0, answer.