The first few values of $n$! are

$$\begin{aligned} & 1!=1 \\ & 2!=2(1)=2 \\ & 3!=3(2)(1)=6 \\ & 4!=4(3)(2)(1)=24 \\ & 5!=5(4)(3)(2)(1)=120 \end{aligned}$$

We note that

$$\begin{aligned} & 2!-1!=1 \\ & 4!-1!=23 \\ & 3!-1!=5 \\ & 5!-1!=119 \end{aligned}$$

This means that if $a$ and $b$ are positive integers with $b>a$, then $1,3,5,9$ are all possible ones (units) digits of $b!-a!$. This means that the only possible answer is choice (D), or 7. To be complete, we explain why 7 cannot be the ones (units) digit of $b!-a!$. For $b!-a!$ to be odd, one of $b!$ and $a!$ is even and one of them is odd. The only odd factorial is 1!, since every other factorial has a factor of 2. Since $b>a$, then if one of $a$ and $b$ is 1, we must have $a=1$. For the ones (units) digit of $b!-1$ to be 7, the ones (units) digit of $b$! must be 8. This is impossible as the first few factorials are shown above and every greater factorial has a ones (units) digit of 0, because it is a multiple of both 2 and 5.