In RSA, a message is encrypted by $m^e \pmod N$. $N$ is the modulus, $m$ is the message and $e$ is the public exponent. (I know that $m$ should not be greater than $N$.)
My question is, can $m^e$ be greater than $N$ (obviously, before taking the modulus)?
In that case is there a possibility like $ m_1^e=m_2^e \pmod N$, i.e. can we get a collision?
Correction to Henricks answer: collisions are impossible (unless someone did something wrong). That is, if:
$e$ is a proper RSA exponent (that is, relatively prime to $p-1$ and $q-1$, where $p$ and $q$ are the factors of $N$), and:
$m_1 \neq m_2 \mod N$ (that is, you're not trying to encrypt the same message twice),
Then we will always have $m_1^e \neq m_2^e \mod N$
This is rather implied by the fact that the RSA operation can be inverted using the decryption exponent; if two different messages collided, then that couldn't be inverted uniquely.
Yes, $m^e$ is in fact supposed to be larger than the public modulus $N$, or else it would be trivial for an attacker with knowledge of nothing but the cipher text and the public exponent to calculate $m$. If $m^e$ is less than $N$, then it is obviously equal to its residue $\bmod N$. Calculating roots is not hard; calculating the root of a residue $\bmod N$ is.
Regarding your second question: As Poncho wrote, as long as the RSA parameters are correctly selected, it is impossible that you will accidentally find two different messages $m_1$ and $m_2$, both greater than 0 and less than $N$, such that $m_1^e \pmod N = m_2^e \pmod N$, because it will only happen if $GCD(e,LCM(p-1,q-1)) \neq 1$.
