Terminology: There are two general "orientations" a path can have:
bottom left to top right; let's call that "slashy"
top left to bottom right; let's call that "backslashy"
These are not mutually exclusive: We'll use the convention that purely horizontal or purely vertical paths belong to both classes.
We'll now prove the claim that given n>0 a covering of an n x n grid by n-1 shortest on-grid paths does not exist.
Otherwise there must be a smallest counter example X. Let n be its size.
We'll show that we can construct from this a counter example Y of size n-1 contradicting minimality of X.
Lemma: If there is a path p in X that contains two distinct points touching two distinct boundaries then Y as above exists.
Wlog p is slashy. Extend p to the bottom left and top right corners (erasing the bits of other paths that are in the way). As this extension hugs the boundaries it will not break up any other path. The resulting configuration X' therefore still has at most n-1 paths. To obtain Y, simply remove (the extended) p and push the resulting two leftovers together.
To conclude the proof of the main claim we will show that a similar extension can always be made even if p only touches one boundary. (Such a p always exists.)
Still wlog p is slashy and starts at the left or bottom boundary. Let y,x be the coordinates where it ends. We now show that we can extend p (and keep doing so until we reach the top or right boundary). If either y+1,x or y,x+1, i.e. the points just above or to the right is part of a slashy path q in X we can extend p along q in the top right direction. If we keep going until the end of q this will not change the total number of paths (at least not upwards, it could in theory reduce it as @Jaap Scherphuis points out). Otherwise both y+1,x or y,x+1 are in backslashy paths. If either is an endpoint we can add it to p. This only leaves the case that y+1,x and y,x+1 are part of the same backslashy path q and connected at y+1,x+1. But then we can simply extend p to y+1,x+1 (which way does not matter). This will temporarily create an additional path because q is cut in two. But q will be rejoined at the final "push-the-two-leftovers-together" stage.