A closet contains 10 pairs of shoes. If 8 shoes are randomly selected, what is the probability that there will be exactly 1 complete pair?

It looks fine to me. In general there are $\binom{10}k\binom{10-k}{8-2k}2^{8-2k}$ ways to choose the $8$ shoes so as to get exactly $k$ pairs, your calculation being the case $k=1$, and a quick numerical check confirms that

$$\sum_{k=0}^4\binom{10}k\binom{10-k}{8-2k}2^{8-2k}=\binom{20}8\;.$$

Here's one way, using permutations.

One pair can be chosen and lined up in ${10\choose 1}\cdot8\cdot7 = 560$ ways and for the remainig shoes, the rest of the numerator ensures that no other pair is selected

$$\begin{align} & \frac{560\cdot 18\cdot 16\cdot 14\cdot 12\cdot 10\cdot 8}{20\cdot19\cdot18\cdot17\cdot16\cdot15\cdot14\cdot13}\\[2ex] \text{which amounts to the same as}\;\; & \dbinom{10}{1}\times \dbinom{9}{6}\times 2^6\Big/\dbinom{20}{8}\end{align}$$

There's another method to solve is problem, though it's pretty long-

There are ${10}\choose{1}$ ways to select a pair. Now each pair has a left shoe and a right. No of ways to choose the remaining 6 shoes such that no pair is selected can be made in the following manner - all left , 5 left 1 right, 4 left 1 right .... till all right shoes are chosen, this becomes- ${9}\choose{6}$ +${9}\choose{5}$${4}\choose{1}$+......

This can then be divided by the ${20}\choose{8}$ ways. The answer turns out to be same as above.