How can one show in detailed steps that Fermionic annihilation and creation operators, under Jordan-Wigner transformation, satisfy the Fermionic commutation relations?
The Fermionic commutation relations are: $$\{\hat{a}_i,\hat{a}_j\}= \{\hat{a}_i^\dagger,\hat{a}_j^\dagger\} =0 , \{\hat{a}_i,\hat{a}_j^\dagger\} = \delta_{ij}.$$
Based on my answer to this: Fermionic occupation operator and nearest neighbor Fermionic hopping interaction as a qubit operator, you can see that we have:
\begin{align} \hat{a}_i &= \frac{1}{2} Z^{\otimes (i-1)} (X - iY),\\ \hat{a}_i^\dagger &=\frac{1}{2} Z^{\otimes (i-1)} (X + iY).\\ \end{align}
If $i=j$ we have:
\begin{align} \{\hat{a}_i,\hat{a}_i^\dagger\} &\propto \frac{1}{4} ((X - iY)(X + iY) + (X + iY)(X - iY)),\\ &= \frac{1}{4} (X^2 + iXY + - iXY + Y^2 + X^2 -iXY + iXY + Y^2) \\ &=\frac{1}{4}(2X^2 + 2Y^2)\\ & =\frac{1}{4}(4I) \\ & = I. \end{align}
All $Z$ operators are replaced by $I$ operators since $Z \times Z = I$, and this is also what was done in the penultimate step, where $X^2 = I$ and $Y^2 = I$ were used.
For the other anti-commutators we have:
\begin{align} \{\hat{a}_i,\hat{a}_j\} &= \frac{1}{4} ((X - iY)(X - iY) + (X - iY)(X - iY)),\\ \{\hat{a}^\dagger_i,\hat{a}^\dagger_j\} &= \frac{1}{4} ((X + iY)(X + iY) + (X + iY)(X + iY)). \end{align}
You can then do the same type of arithmetic that I went through in detail for the first anti-commutator.