Let's start from the basics. Any arbitrary single qubit state can be written as

$$|\Psi\rangle = e^{i\gamma} \left(\cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle\right),$$ where $\theta, \phi, \gamma \in \Bbb R$. $0\leq \theta \leq \pi$ and $0\leq \phi\leq 2\pi$. $e^{i\gamma}$ is a global phase here. Qubit states with arbitrary values of $\gamma$ represent the same quantum state as global phases don't have any *observable effects*. So the standard notation for qubit states is

$$|\Psi\rangle = \cos \frac {\theta}{2} |0\rangle+ e^{i\phi} \sin \frac{\theta}{2}|1\rangle $$

where we neglect the global phase factor $e^{i\gamma}$.

Remember that you are working on a vector space structure where $|0\rangle$ and $|1\rangle$ are your basis elements. The rotation operators $R_X, R_Y$ and $R_Z$ are simply linear maps in this context (in the standard $\{|0\rangle, |1\rangle\}$ basis). And so are the Pauli $X, Y$ and $Z$.

Now, note that you can write $R_X(\pi)$ as $-iX$. When $X$ acts on an arbitrary state $|\Psi\rangle$ it will produce $X|\Psi\rangle$. If $-iX$ acts, it will produce $-iX|\Psi\rangle$ instead. However, the states $-iX|\Psi\rangle$ and $X|\Psi\rangle$ simply differ by a global phase of $-i$, or in other words by a factor of $e^{i\gamma}$ where $\gamma = (2n+1)\frac{\pi}{2}$ and $n\in \Bbb N$. So, as you can see, $R_X(\pi)$ and $X$ are equivalent mappings as global phases don't matter in quantum mechanics!

As for *why* the Pauli operators and rotation operators differ by a phase factor, you'll have to check out the derivations for the following.
You'll find an outline here.

$$R_X(\theta) = e^{-i \frac{\theta}{2}X},$$

$$R_Y(\theta) = e^{-i \frac{\theta}{2}Y},$$

$$R_Z(\theta) = e^{-i \frac{\theta}{2}Z}.$$