Odd last digit

Once $5$ gets into the last digit there's no way to get it out of there.

The expression $1 \times 3 \times 5 \times 7 \times \dots \times 97 \times 99$ is odd (since it is the product of all odd numbers) and divisible by $5$ (since $5$ is one of its factors).

Since every odd number divisible by $5$ must end in a $5$ , the last digit of the answer is $\boxed{5}$ .

If $n> 3$ is an odd number then $\begin{aligned} n!! & = n \,(n-2)\,(n-4) \,(n-6)\times \cdots \times 5 \times 3\times 1 \\ &= 5 \,(2k+1) \mod(10) = 5\end{aligned}$ Same problems here and here