How do we know that entropy is a state function given by $\delta Q_{\mathrm{rev}}/T$ for any arbitrary (not just reversible process)? In this video, the professor goes through a "derivation" for entropy using the Carnot cycle, in which he establishes that entropy is a state function for that process.
However, he then continues the lecture and then generalizes that entropy is a state function for any process. (e.x. around 49 minutes). The essence of his argument is that since entropy is a state function, we can calculate its change between any 2 points using a reversible process, even if it happened irreversibly in real life.  However, my issue is, how do we know it is a state function?  We were only able to say it is a state function when we went through a derivation using the Carnot cycle (a reversible process).  It doesn't seem justified to say that is fact holds for any arbitrary process.
https://www.youtube.com/watch?v=ouSLRgkPzbI
To summarize, I am confused by the fact that our discovery that entropy is a state function was only arrived at when we looked at the Carnot cycle, a reversible process.  However, how do we know that entropy is also a state function given by $\delta Q_{\mathrm{rev}}/T$ for any arbitrary (e.x. irreversible process)?  
 A: 
However, my issue is, how do we know it is a state function? We were only able to say it is a state function when we went through a derivation using the Carnot cycle (a reversible process). It doesn't seem justified to say that is fact holds for any arbitrary process.

Entropy is usually thought of mathematically as a function of thermodynamic state variables.
"Entropy is state function" means that for given thermodynamic system (say gas in a cylinder with movable piston), all it suffices to determine value of entropy are values of all its state variables (say, internal energy and volume of the gas).
But this is indeed not clear from the original definition of entropy: entropy of state $X$ is often written or implied to be
$$
S(\mathbf X) = \int_{\mathbf X_0}^{\mathbf X} \frac{dQ}{T}
$$
where $dQ$ is infinitesimal element of heat absorbed by the system and the integral is calculated along any continuous path connecting $X_0$ with $X$.
And you are right this does not make it clear why the expression does not depend on the kind of process that takes the state from $\mathbf X_0$ to $\mathbf X$.
The traditional way to show that the path does not matter and so the expression is a function of final $X$ only is based on the theoretical result that is true for the Carnot cycle with ideal gas. The result is that in such a cycle, sum of reduced heats is zero:
$$
\frac{\Delta Q_1}{T_1} + \frac{\Delta Q_2}{T_2}  = 0.
$$
This is then used to prove that for any reversible process that ends up at the same state it began, the integral along the closed path
$$
\oint \frac{dQ}{T} = 0.
$$
This holds irrespective of the details of the path such as its shape or length. The usual way to do this proof is to replace the actual closed path (that is otherwise of arbitrary shape) by an alternating sequence of very tiny isotherms and adiabats that go very near the actual path. Those isotherms and adiabats can be thought of as part of fictive Carnot cycles and it can be shown that sum of reduced heats of those Carnot cycles approximates the above integral. But the sum of the reduced heats is 0 for those Carnot cycles. The smaller the Carnot cycles, the better the approximation to the original path, in the limit of zero size the isotherms and adiabats on the outer edge become the same as the original path.
After that is done, a mathematical theorem from multivariable calculus is used: if the loop integral is zero for any loop, the expression
$$
\int_{X_0}^{X} \frac{dQ}{T}
$$
is a function of the points $X_0,X$ only, the actual path along which the integral is evaluated does not matter. One point ($X_0$) is usually set to be fixed and then the function is function of the final point ($X$) only.
