Is it possible to reach the initial arrangement? 
We have a stack of $n$ books piled on each other, and labeled by $1, 2, ..., n$. In each round we make $n$ moves in the following manner: In the $i$-th move of each turn, we turn over the $i$ books at the top, as a single book. After each round we start a new round similar to the previous one. Show that after some moves, we will reach the initial arrangement.


Say $n=4$ and initial arrangement of books $(a,b,c,d)$. First we act on it with identical transformation $\pi_1=id$ which leaves everything as it was. Then we act on it with 
$$\pi_2 = \left(
\begin{array}\\
   1 & 2 & 3 & 4  \\
    2 & 1 & 3 & 4 
\end{array}\right)$$
and we get $(b,a,c,d)$, then we act on this one with $$\pi_3 = \left(
\begin{array}\\
    1 & 2 & 3 & 4  \\
    3 & 2 & 1 & 4 
\end{array}\right)$$ and we get $(c,a,b,d)$ and then $$\pi_4 = \left(
\begin{array}\\
   1 & 2 & 3 & 4  \\
    4 & 3 & 2 & 1 
\end{array}\right)$$ and we get $(d,b,a,c)$ and then we repeat acting with $$\pi_1,\pi_2,\pi_3,\pi_4,\pi_1,\pi_2,...$$
Now what do we get if we repeat enough time $$\sigma = \pi_4\circ \pi_3\circ \pi_2\circ \pi_1$$ on starting $(a,b,c,d)$? If we repeat this $\sigma $ exactly $24$ times (which is the order of symmetric group $S_4$) we shoud get initialy arrangment. Clearly this can be easly generalized for arbitrary $n$. 
Is this correct?

Edit: As suggested in comment by Lord Shark the Unknown, It shoud be considered also the first and last front of a book. So I should observe  8-couple $(a1,a2,b1,b2,c1,c2,d1,d2)$, where $x1$ is first front and $x2$ last one, instead of 4-couple $(a,b,c,d)$ and act on it with $S_8$?
 A: Let $S$ be the set of $2n$ book covers. Each round is a permutation of $S$. Every permutation $\pi$ of a finite set has a number $k$ such that $\pi^k=\text{id}$. One can choose $k$ to be the lcm of the cycle lengths of $\pi$. 
For example, when $n=2$, a single round looks like this, where capital letters are the top cover and lower case are the bottom cover. 
A    a    b
a    A    B
B    B    A
b    b    a

In cycle notation, this looks like $(A\;\; B\;\; a\;\; b)$. This a cycle of order four, so four rounds suffice to return the books to their original order. 
A: For clear reference, here is a complete cycle of moves. Negative number represents book face down.
 1  2  3  4 
-1  2  3  4 
-2  1  3  4 
-3 -1  2  4 
-4 -2  1  3 
 4 -2  1  3 
 2 -4  1  3 
-1  4 -2  3 
-3  2 -4  1 
 3  2 -4  1 
-2 -3 -4  1 
 4  3  2  1 
-1 -2 -3 -4 
 1 -2 -3 -4 
 2 -1 -3 -4 
 3  1 -2 -4 
 4  2 -1 -3 
-4  2 -1 -3 
-2  4 -1 -3 
 1 -4  2 -3 
 3 -2  4 -1 
-3 -2  4 -1 
 2  3  4 -1 
-4 -3 -2 -1 
 1  2  3  4 << return to initial state

