Understanding modulus math for RSA I probably have this doubt due to poor understanding of modular arithmetic, so an explanation would be greatly appreciated.

In any standard explanation of RSA, the following is present:

c = m^e mod n (where, c is the cipher text, m is the message, e the public key exponent, and n is the modulus)

And for decryption:

m = c^d mod n

To prove this, I've seen that the next step normally shown is :

m^(e.d) = m mod n

My question is, when c is raised to the private key exponent, why is the exponent not distributed over both m and mod n? I.E why is it not m^(e.d) mod n^(e.d)
You have to understand how the modulus operation works. When we say $a=b\mod c$, what we mean is that $a-b$ is a multiple of $c$. Now, since for any positive integer $n$, $a^n-b^n$ is a multiple of $a-b$, we can usually raise both sides to the same power in a modular equation, keeping $c$ intact. However, if $a-b$ is a multiple of $c$, there is no guarantee that $a^n-b^n$ is a multiple of $c^n$. Hence, raising $c$ to the same power makes no sense. I hope this answers your question. For a better understanding of elementary number theory, you can go through the initial chapters of a high-school text (like Burton's).