Attached my solution

Minor correction: In your definition of , you wrote , and it should have been .


Also, while it's fairly obvious, you probably should have noted somewhere along the way that any symmetric spinor is completely defined by the components .

These happen to be the same components that appear (individually) in the coefficients of , which is what allows us to go from equating the polynomials and directly to the required spinor identity.

That said, you could then simplify the last bit of your proof, from (**) down, by noting that




and thus






which, given my preceding comment, establishes immediately that , since they are both symmetric spinors that yeild the same polynomial when contracted with .

There's another omission from the attached proof, which is to note that there are special case(s) when is of rank < n, which can happen if ( rank n-1), or ( rank n-2),... etc.

Assuming the actual rank of is r, we can deal with this by redefining (i.e. defining differently than given in the current proof) only the first (n-r) spinors from the list as:



Here is the coefficient of the highest rank term in .
With this modification, the rest of the proof continues to hold.





Note that this addition to the proof is also essential for understanding how the standard Majorana basis states work, because these states are composed of symmetrized products of the single-index spinors and , the latter of which just doesn't make sense in light of the proof as originally given!

Yes, I forgot the case of polynomials of rank less than n, that of course is a possible physical situation.
So your remark is necessary to cmplete the proof.