Is knowing the private key of RSA equivalent to the factorization of $N$? Given the RSA modulus $N$ the fastest method to factor it is of sub-exponent order. But, now if I know the private key $d$ of RSA, does that mean I can factor $N$ efficiently?. It intuitively seems true as I know the public/private key pair. Can someone help me to prove this?

Please provide references so that I can read further into this subject.
Yes, they are (deterministically) equivalent. The original RSA paper (Section IX.C), working off Miller's results (Theorem 3), showed how knowing the secret exponent $d$ was probabilistically equivalent to factoring $n$.

Later, using more advanced techniques, Coron and May showed how to deterministically reduce finding $d$ to factoring $n$.