The answer by Stefanos is better, since it puts together two natural ingredients. There are other ways. Imagine instead putting the books on a single shelf, together with $6$ books dividers. (Then for the real shelving, we will put the books up to the first divider onto Shelf $1$, and the books from the first divider to the second onto shelf $2$, and so on.)
Imagine that the books are called $a,b,c,d,e$. Write $D$ for a divider. We want to count the number of $11$-letter "words" that use each of $a,b,c,d,e$ once, and $6$ copies of $D$.
If the $Ds$ were distinct, say $D_1$ to $D_6$, there would be $11!$ words. Since they are identical, we divide by $6!$.