Let $P:C^{\infty}(M) \to C^{\infty}(M)$ be a pseudo differential operator of order $d$ on a compact manifold $M.$
Then $P$ is said to be elliptic if there exists an operator $Q:C^{\infty}(M) \to C^{\infty}(M)$ of order $-d$ such that $PQ-I$ and $QP-I$ are infinitely smoothing operators. Now since $PQ-I$ is infinitely smoothing the map can be extended to $$PQ-I:H_s(M) \to H_{s+1}(M)$$ here $H_s(M)$ denotes the sobolev spaces for a real $s$. Now since the manifold is compact the inclusion $i:H_{s+1}(M) \to H_{s}(M)$ is a compact operator. So the map $$PQ-I:H_s(M) \to H_{s}(M)$$ is compact. Similar argument can be used to show $$QP-I:H_s(M) \to H_{s}(M)$$ is compact. This implies that the operator $$P:H_s(M) \to H_{s-d}(M)$$ is Fredholm. Here the definition of Fredholm operator I am using is that it has to invertible modulo compact operator. So is this argument right?
The reason I suspect I am missing something is because in the argument to show $PQ-I$ is compact I only used the fact that it is of order $-1$. But in all the references I checked they seem to say that infinitely smoothing operators are compact. So am I missing something? Thank you.
The fact that the remainder is smoothing is used to say that the index of $P$ is independent of $s$, since its kernel and cokernel consists of smooth functions. Not only smoothing operators are compact, if the order of $P$ is less than $d$ then, as a map $H_s\to H_{s-d}$, $P$ is compact. I would say that $P$ is elliptic when its leading symbol is invertible away from zero, and then your definition is a consequence.