RSA-CRT exponent reduction In the implementation of RSA-CRT, the exponent d is reduced mod p-1 ($d_p = d \bmod {(p-1)}$). The only proof I've found for that, is the following (considering $d = k\varphi(p) + d \bmod {\varphi(p)}$:
$$c^d = c^{k\varphi(n) + d \bmod {\varphi(p)}}$$
$$(c^{\varphi(p)})^k * c^{d\bmod {\varphi(p)}} \equiv (1)^k * c^{d\bmod {(p-1)}} (\bmod p)$$
Source: https://www.di-mgt.com.au/crt_rsa.html
So $c^d \equiv c^{d(\bmod(p-1))} (\bmod(p))$ However, I can't see why is the Euler theorem valid here, as c is not granted to be coprime with p (as far as I know). Any hint on this?
You're correct, the proof isn't precisely correct, because we don't necessarily have $c^{\phi(n)} = 1$, specifically in the case $c \equiv 0 \pmod p$.

Here is a more correct approach; we have $c^1 \equiv c \pmod p$ (trivially), and $c^{p-1} \equiv c \pmod p$ for any $c$, prime $p$ (Fermat's little theorem [1]).  By induction, we get $c^{k (p-1) + 1} \equiv c^1 \pmod p$ (for any $k$), and hence $c^{k (p-1) + \ell} \equiv c^\ell \pmod p$.

If we designate $x = k(p-1) + \ell$ for $0 \le \ell &lt; p-1$ (and any $x$ can be put in that form), we have $c^x \equiv c^{x \bmod p-1} \pmod p$, for any $x, c$ and prime $p$; in particular, if $x = d$



[1]: The most common expression of Fermat's Little Theorom is that $1 \equiv x^{p-1} \bmod p$ for prime $p$, $x$ relatively prime to $p$.  However, an equivalent formulation (and what Fermat originally stated) is that $x \equiv x^p \bmod p$ for prime $p$, any $x$ - I'm using this alternative formulation.