Do Not Multiply Out

Notice that $\text{last digit of } n!! = 5 , \text{if n is odd number > 3 }$ Since $9$ is an odd number. Thus, the last digit of $9!! = 1 \times 3 \times 5 \times 7 \times 9$ is $\boxed{5}$ .

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If $n> 3$ is an odd number. Then $\begin{aligned}n!! & = n \times (n-2) \times (n-4) \times \cdots \times 5 \times 3 \\ & = 5 \left\{ \underbrace{n \times (n-2) \times (n-4) \times \cdots \times 3}_{\text{odd number }}\right\} \\ & = 5\,(2k+1)\mod(10) =5\phantom{c} , k \in\mathbb N \end{aligned}$ Thus the last digit of $n!!$ is always $5$ for $n>3$ .

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Problem After being inspired by this

The result is a multiple of 5,so it ends in 5 or 0,but since it is an odd number,the answer should be 5.

Since we are multiplying by $5$ and that there are no even numbers, the units digit will be $\boxed{5}$ .