Among 4 consecutive integers there always exist two even numbers, one of which is also divisible by 4. Thus the quotient is even, excluding all odd final digits.
If the final digit of the quotient upon division by 4 is 0, 2, 4, 6 or 8, then the
final digit of the dividend is 0, 8, 6, 4 or 2, respectively.

* If the four consecutive integers have final digits 1, 2, 3 and 4, then the final digit of their product is 4.
* If the four consecutive integers have final digits 6, 7, 8 and 9, then the final digit of their product is 4.
* In all other cases one of the four integers is divisible by 5. Since the product is even, its final digit can only be 0.

So the product can have a final digit of 4 or 0. Upon division by 4, the final digit of the quotient is 6 or 0, respectively.

Among 4 consecutive integers there always exist two even numbers, one of which is also divisible by 4. Thus the product is divisible by 8. We consider two cases.

* If one of the factors is divisible by 5, then the product is also divisible by 5, and thus also by 40. Therefore the quotient upon division by 4 is divisible by 10 and thus ends with a 0.

* If none of the factors are divisible by 5, then they are of the form $5n + 1, 5n + 2, 5n + 3$ and $5n + 4$. The expression $(5n + 1)(5n + 2)(5n + 3)(5n + 4)$ can be written as $5m + 1 \cdot 2 \cdot 3 \cdot 4$ for an integer $m$. As the product of the four numbers and the term $1 \cdot 2 \cdot 3 \cdot 4$ are divisible by 8, then so must also be $5m$ and thus also $m$. Therefore the product is of the form $40k + 24$, which yields $10k + 6$ upon division by 4. This number ends with a 6.