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, $$c^{e^{k-1}} = c^d = m \pmod{n}.$$
Let us remind that, by definition of the RSA encryption, we have $c = m^e \bmod{n}$ (where $n=pq$ and $\mathrm{gcd}(e, (p-1)(q-1)) = 1$, but it's not important here). Let's take the equation $$c^{e^{k-1}} \equiv m \bmod{n}$$ and let's raise both sides to the power $e$: $$\left(c^{e^{k-1}}\right)^e \equiv m^e \bmod{n}\,,$$ so $$c^{e^k} \equiv c \bmod{n}\,.$$
