If it is true prove that
for every $a,m$ ($a,m$ are co-primes), there exists $a^{-1}$ such that $a\cdot a^{-1} \equiv 1\bmod{m}$, which can be found by $$a^{\phi(m)-1}$$ where $\phi(m)$ is the totient function.
I need a proof of this statement that $a^{t(m)-1}$ is mod inverse of $a$.
You may also prove $a^{\phi(m)} ≡ 1 \mod m$.
This is true: $a^{\phi(m)}\equiv1\pmod m$, when gcd($a,m)=1$, and hence the modular inverse for $a$ is $a^{\phi(m)-1}$. This is an old theorem, (more than 250 years ago) due to Euler and is found in all textbooks on elementary number theory, along with Fermat's Little Theorem. This is a conceptual fact. However, for large numbers this is not a practical way to compute the modular inverse: because this presupposes factorization of $m$ is known. At present there is no formula for $\phi(m)$ without knowing the factorization of $m$.
In fact the security of the RSA cryptosystem rests on the practical difficulty of factorizing large numbers.
The usual way for finding the modular inverse is carrying out Euclid's algorithm for gcd with extra details (keep track of quotients in every division, not just remainders). This is called Extended Euclidean Algorithm.