Is it true that $(c_1c_2*...*c_k)^m = c_1^mc_2^m*...*c_k^m$ if every $c_i$ is a cycle and every pair $(c_i,c_k)$ is disjoint without any other condition?
I have seen some proofs that to my understanding use this fact but I cannot find it expicitly somewhere written and that's usually a sign that does not hold. But I tried to prove it for cycles of length 2 and seems like it works. I could not find it though, if it is a duplicate refer me to the corresponding question but please let it be something that verifies my statement and not some close to it because I am new to the field of abstract algebra.
Yes. You don't even need to know that they are cycles, only that they commute (for example, their supports are disjoint).
In general, if $G$ is an arbitrary (semi)group and $c_1,\ldots,c_k\in G$ commute, then straightforward induction shows that $(c_1c_2\cdots c_k)^n=c_1^nc_2^n\cdots c_k^n$.