Let $p,q$ are primes and $n = pq$ as in every RSA setting and now use a random $e$ that holds the following properties
- $gcd(e, \phi(n)) \neq 1$
- $(e^{-1} ~\text{mod} ~\phi(n))^{4}\cdot3 < n$
- $e^{-1} ~\text{mod} ~\phi(n) < \sqrt[3]{n}$ (integer square root), where $\sqrt[3]{n} \in \mathbb{Z}$
where $\phi$ is euler's totient function. This $e$ is used as the public exponent for the public key and $d$ is the private exponent for the private key.
Remember that $\phi(n) | ed - 1$, as $ed = 1 + k \cdot \phi(n)$ holds for $k \in \mathbb{Z}$. The question is, why it holds that $$(e^{-1} ~\text{mod}~ n \cdot \phi(n)) ~\text{mod}~ \ \phi(n) = e^{-1} ~\text{mod}~ \phi(n)\text{?}$$ Could someone explain it mathematically or give a proof why that holds?
A related question regarding multiple of $\phi(n)$ is asked in this question. Unfortunately, I don't understand the relation between the multiple of $\phi(n)$, the $gcd$ and why this equation $ed = 1 ~\text{mod}~ k \cdot \phi(n)$ holds $ed = 1 ~\text{mod}~ \phi(n)$ for $k \in \mathbb{Z}$.
Additionally it would be nice to read a proof for the related question regarding the multiple of $\phi(n)$, if someone knows why that holds.