Why does the action appear in the propagator in the path integral formulation of quantum mechanics?

Question

According to modern quantum mechanics by Sakuria, Feynman's path integral formulation of classical mechanics was motivated by a statement in Dirac's book: $$exp\left(i\int^{t_{2}}_{t_{1}}\frac{dt\;L_{classical}(x,\dot{x})}{\hbar}\right)\;\text{corresponds to}\;\langle x_{2},t_{2}|x_{1},t_{1}\rangle$$ where $\langle x_{2},t_{2}|x_{1},t_{1}\rangle$ can be interpreted as the probability amplitude for a particle prepared at $t_{1}$ with position eigenvalue $x_{1}$ to be found at later time $t_{2}$ at position $x_{2}$. I can't find any justification for this expression in any of the sources I've read on the path integral formulation of quantum mechanics. The closest I've come was a paper written by Dirac (https://www.hep.anl.gov/czachos/soysoy/Dirac33.pdf), however I struggled to make sense of it. I feel comfortable with path integral formulation taking this as a given, so my question is why does the action appear in an exponential in this way to represent the transition amplitude?