How does the cyclic attack on RSA work? I am trying to get the idea of cyclic attacks againts assymetric RSA encryption.
Taken from Handbook of applied cryptography .

Let $k$ be a positive integer such that $$c^{(e^{k})} = c\mod n  \tag{1}.$$
There for
$k-1$ it holds that $$c^{(e^{k-1})} = m \mod n  \tag 2$$ where $m$ is the message for encryption $n$ is the modulus and $c$ is the ciphertext.

I can't understand why equation (2) must hold?
We start with the definition of textbook RSA encryption: $c = m^e \bmod n$. From your first equation

$$c^{e^k} = c \pmod{n},$$ we have that if $c^{e^k} = c \pmod{n}$, then $e^k = 1 \pmod{\phi(n)}$ (Euler's theorem). Dividing both sides by $e$, we get 

$$e^{k-1} = e^{-1} \pmod{\phi(n)}.$$ 

By definition, $d = e^{-1} \pmod{\phi(n)}$. Thus, $$c^{e^{k-1}} = c^d = m \pmod{n}.$$