The mod $p$ Galois representation of the Frey curve is unramified away from $2, p$ Given a hypothetical solution to Fermat's last theorem for $p \ge 5$ $$a^p + b^p + c^p = 0$$with $a \equiv -1 \pmod 4$, $b$ even, we can write down the Frey Curve$$E: y^2 = x(x-a^p)(x+b^p)$$which has discriminant $\Delta_E = -2^{-8}(abc)^{2p}$.
Letting $G_\mathbb Q =\mathrm{Gal}(\overline{\mathbb Q}/\mathbb Q)$ act on the $p$-torsion points of $E$, we obtain a Galois representation $$\rho:G_\mathbb Q \to \mathrm{Aut}(E[p])\cong \mathrm{GL}_2(\mathbb F_p)$$
I've seen in numerous texts that

$\rho$ is unramified at all primes $\ell \ne 2, p$

but most of these texts just say that this is "easy to check".
Is there an easy way to see why this is true? I'm guessing I'm missing something obvious!
 A: Well, it was easy for Serre to check ;D
This is not obvious, but once you know the right tools to use, it does become an easy exercise. A great resource for learning about the proof of FLT is this expository paper by Darmon, Diamond, and Taylor: https://www.math.wisc.edu/~boston/ddt.pdf
In particular, see Theorem 2.15 (b). The proof refers you to other non-trivial propositions, but you'll see that, with the relevant references in hand, you can indeed check that the representation is unramified away from $2,p$. To make this a worthy post, I'll sketch the idea below.
You've set things up so that the discriminant of $E$ is $\Delta = 2^{-8}(abc)^{2p}$. For any elliptic curve, the only primes at which ramification is possible are those which divide the discriminant. In particular, for the Frey curve, the only primes at which $\rho$ can ramify are $2,p$, and the primes $\ell$ dividing $abc$.
Our situation is even better, because this discriminant is minimal. In fact, if $\ell \mid abc$, then $\rho$ is unramified at $\ell$ if and only if $p \mid v_{\ell}(\Delta)$ (where $v_\ell$ is the $\ell$-adic valuation). This is Proposition 2.12(c) at the link I provided. So in fact $\rho$ is unramified at all those primes $\ell$ which divide $abc$. 
So the only primes left at which our representation can ramify are $2$ and $p$.
A: To elaborate a little more on the existing answer:
There is a criterion (Neron--Ogg--Shafarevic) whichs says that $E$ has good reduction at $\ell \neq p$ iff the Galois action on the $p$-adic Tate module is unramified.
So if $\ell$ divides the (minimal) discriminant, and so is a prime of bad reduction, then we know that the Galois action on the $p$-adic Tate module will be ramified.
Recall that the $p$-adic Tate module is built as the inverse limit of $E[p^n]$ over all $n$.  This suggests the more subtle question, namely: how large a value of $n$ do we have to take in order to start seeing ramification at $\ell$.
It turns out that, in the case of semi-stable reduction, this is related to the power of $\ell$ that divides the discriminant: for example, the value $n =1$ is not large enough (i.e. the Galois action on the $p$-torsion is unramified at $\ell$, despite $\ell$ being a prime of bad reduction) iff the power of $\ell$ dividing the discriminant is a multiple of $p$.
The usual way to see this is via the theory of Tate curves.  The rough idea is that having semistable reduction means that the elliptic curve becomes a nodal curve mod $\ell$, and the group law on the nonsingular points of a nodal cubic is isomorphic to the multiplicative group $\mathbb G_m$.  The $p^n$-torsion on here is equal to the $p^n$th roots of $1$, and hence has size $p^n$, rather than $p^{2n}$ (the size of $E[p^{2n}]$).  Thus a lot of the $p^n$-torsion on $E$ ``disappears'' somehow upon reduction mod $\ell$, and this is reflect in the appearance of ramification.    
However, if the power of $\ell$ in the discriminant is a multiple of $p$, then the reduction of $E$ at $\ell$ is not actually a nodal curve --- rather, it is the union of $p$ lines, each meeting at a point, arranged in a $p$-gon (actually a $p^r$-gon if $p^r$ is the power of $\ell$ dividing $\Delta,$ but let me write as if $r = 1$).  (To see the relationship with a nodal curve, think of a nodal curve as a single line meeting itself in a point --- so a $1$-gon.)  If we remove the singular points, we get not just $\mathbb G_m$, but $\mathbb G_m \times (\mathbb Z/p)$, and this has $p^2$ $p$-torsion points; thus there is ``enough room'' for all the $p$-torsion on $E$ to reduce nicely mod $\ell$, and so we don't get any ramification at $\ell$ on the $p$-torsion.
