Why is Euclidean Time Periodic?

I don't think that Wick rotated time $\tau$ is periodic by itself. But it turns out that thermal averages of operators are periodic with respect to the variable $\tau$. Consider a generic time dependent operator $\hat{A}(\tau)$ with the standard time evolution expansion $\hat{A}(\tau) = e^{\hat{H}\tau} \hat{A}(0) e^{-\hat{H}\tau}$ and consider its thermal average $A(\tau) \equiv \hat{\left\langle A (\tau) \right\rangle } = Z^{-1} \mathrm{Tr}[e^{-\beta \hat{H} }\hat{A}(\tau)]$, where $Z$ is the parition function. You can prove rather simply that $A(\tau + \beta) = A(\tau)$ by exploiting firstly the fact that $ e^{-\beta\hat{H}} e^{\beta\hat{H}} = 1$ and secondly the cyclic property of the trace (I'll leave this as an exercise).

However, not all the objects that we are interested in are necessarily periodic. A remarkable example is the Green function at positive time $\tau \geq 0$ $$ G_{kp}(\tau) = - \left\langle \hat{\psi}_k(\tau) \hat{\psi}_p^{\dagger}(0) \right\rangle $$ which is written in terms of time dependent field operators. In fact you can prove that $G_{kp}(\tau+\beta) = \zeta G_{kp}(\tau)$, where $\zeta = +1$ if $\hat{\psi}$ is a bosonic operator, and $\zeta = -1$ if it is fermionic, so that the function is either periodic or antiperiodic.

In conclusion, the (anti)periodicity of functions with respect to euclidean time relies on how you compute thermal averages.

If you make the substitution: $$ \frac{\mathrm{i}t}{\hbar}\rightarrow\frac{1}{k_{\mathrm{B}}T} = \beta, $$ the quantum evolution operator becomes: $$\mathrm{e}^{-\mathrm{i}\frac{\hat{H}t}{\hbar}} \rightarrow \mathrm{e}^{-\beta \hat{H}},$$ which looks familiar in the context of thermal statistical field theory since the partition function $Z$ is given by: $$ Z = \mathrm{Tr}\left [ \mathrm{e}^{-\beta \hat{H}} \right ].$$

$\tau$ is related to $\mathrm{i}t$ with maybe a minus sign and some conventions on units.

If you think of $\mathrm{e}^{-\beta \hat{H}}$ as an evolution operator $\mathrm{e}^{-\tau \hat H}$, taking a state and evolving it from $\tau = 0$ to $\tau = \beta$, then it's clear that the boundary conditions are periodic.

Because if you take $\tau = 2\beta$, then: $$ \mathrm{e}^{-2\beta \hat{H}} = \mathrm{e}^{-\beta \hat{H}}\cdot \mathrm{e}^{-\beta \hat{H}}, $$ i.e. the first "evolution operator" evolves from $\tau = 0$ to $\tau = \beta$, and the second one also. I.e. the end-time of the first evolution operator becomes the new $\tau = 0$ for the second evolution operator. So the evolution is periodic. In the context of thermal statistical field theory. Don't know about generic Euclidean time.