Why, intuitively, is Lob's theorem true? Lob's theorem states that:
Let $\textbf{Prov}(n^A)$ be the arithmetic statement such that $PA\vdash$ $A$ iff $\textbf{Prov}(n^A)$, where $PA$ is peano arithmetic, and $n^A$ is the godel number of $A$.
Then 

$$PA \vdash\textbf{Prov}(n^A)\rightarrow A\quad\quad\text{
implies} \quad \quad PA\vdash A$$

Where the same applies to any system that is at least as powerful as peano arithmetic.
My question is: Is there an intuitive explanation of why this is true? I've read an $n$ step proof in modal logic, but my intuition is not improved by it.
 A: It's just a reformulation of Gödel's theorem : if you have some intuition for Gödel's theorem then you can transfer it to Löb's theorem.
Indeed if PA proves $Prov(A)\to A$ then it means that the provability of $A$ is enough to conclude the truth of $A$ : but we know that some models may believe that some things are provable while they're not : this is Gödel's theorem with $A= "0=1"$. 
So if $Prov(A)\to A$ is provable it means that any model that thinks it has a proof of $A$ is not wrong about it : this can only happen if $A$ is true. 
To see more precisely why it's just Gödel's theorem : assume you have such an $A$, and assume $T=$PA + $\neg A$ is consistent. In particular $T$ is a consistent recursively axiomatizable extension of PA, so it doesn't prove its own consistency (Gödel); in particular there is no $\varphi$ such that $T\vdash \neg Prov_T(\varphi)$.
However, $PA\vdash Prov_{PA}(A)\to A$ so $T\vdash \neg Prov_{PA}(A)$ hence $T\vdash \neg Prov_T(0=1)$ and so $T\vdash Con(T)$ : a contradiction.
Conversely, Löb's theorem can be used to prove Gödel's theorem for PA: if PA is consistent, then PA$\nvdash 0=1$ and so by Löb PA$\nvdash Prov_{PA}(0=1) \to (0=1)$ hence  PA $\nvdash \neg Prov_{PA}(0=1)$: PA doesn't prove its own consistency (this of course works for any nice $T$ extending PA)
Addendum : I've reduced "intuition for Löb's theorem" to "intuition for Gödel's theorem" but haven't given the latter. For this one it's "simply" that a theory satisfying the hypotheses can actually talk about itself and its proofs and so can in effect reproduce the liar's paradox : there is a sentence that represents "I am not provable"
