Could someone please elaborate on how $(c^d \bmod n) \bmod p = c^d \bmod p$, given that $n = pq$, where $p$ and $q$ are prime numbers?
Well, lets see if we can go through it from the basics.
$x \bmod y$ is the unique integer $x + \ell y$ that satisfies $0 \le x + \ell y &lt; y$, for some integer $\ell$ (which might be positive, negative or zero)  I will skip the parts that prove that, if $x, y$ are both integers and $y &gt; 0$, then there exists such an $\ell$ and it is, in fact, unique.
Now, when we have $(x \bmod pq) \bmod p$, we can replace the inner $(x \bmod pq)$ with $x + \ell_1 pq$, we get $x + \ell_1 pq \bmod p$.  This in turn becomes $x + \ell_1 pq + \ell_2 p$.  We further note that $\ell_4 = \ell_1 q + \ell_2$ is an integer (which we will denote at $\ell_4$) [1], and so we have:
$$(x \bmod pq) \bmod p = x + \ell_4 p$$
On the other side, $x \bmod p$ is the same as $x + \ell_3 p$, that is,
$$x \bmod p = x + \ell_3 p$$
So, on one side, we have $x + \ell_4 p$, which is a value between 0 and $p-1$.
On the other side, we have $x + \ell_3 p$, which is also a value between 0 and $p-1$.
Since both $\ell_3$ and $\ell_4$ are integers, these must be the same, and hence the original values $(x \bmod pq) \bmod p$ and $x \bmod p$ must also be the same.
Then, replace $x$ with $c^d$ and $pq$ with $n$, and there you go...
[1]: BTW: this is the step where we assume that $p$ is a factor of $n$ if that is not true, then relation will not hold in general.