Counting heptagons formed using vertices of $n$-sided polygon having no sides in common with the polygon

I take it that the heptagons are convex.

Let us denote the $7$ chosen vertices as $\Large\circ$ and the rest as $\Large\bullet$

Looking clockwise make blocks of used-unused vertices $\boxed{\Large\circ\Large\bullet},$ treating each as one object.
This ensures that used vertices can never be adjacent,
and hence no side of the polygon is used in forming the heptagon.

There are now $7 + n -14 = n-7$ objects, and the blocks can be placed in $\binom{n-7}{7}$ ways,

but we are giving the blocks only $(n-7)$ starting points instead of $n$,
so we need to use a multiplication factor of $\dfrac{n}{n-7}$,

giving the answer $\dfrac{n}{n-7}\times\dbinom{n-7}{7}$

Another way

Between the $7$ blocks, there are $7$ "compartments" in which the remaining $(n-14)$ can be placed, in $\binom{n-14+7-1}{7-1} = \binom{n-8}{6}$ ways.

But due to the circular type of arrangement, each pattern will repeat $7$ times,
thus ans = $\dfrac{n}{7}\dbinom{n-8}{6}$ ways

Check that the two answers are equivalent !

Let us count the number of such heptagons where one of the vertices is called the leader.

A priori we can choose the leader in $n$ ways. After the leader has been chosen an admissible heptagon can be encoded as a word $$L\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ 1\ \underline{\quad}\ L$$ with $6$ ones and at least one zero in each of the $7$ slots. This leaves $n-14$ zeros to distribute freely in the $7$ slots, which can be done in ${n-14+6\choose 6}$ ways. It follows that there are $$n\cdot{n-8\choose 6}$$ admissible heptagons with a leading vertex.

In this way we have counted each "blank" heptagon $7$ times, so that the number of "blank" heptagons comes to $${n\over 7}\cdot{n-8\choose 6}\ ,$$ as obtained also by true blue anil.