Is 1|V⟩ = |V⟩ provable from Shankar's vector space axioms?

[AxiomOfChoice]

On pg. 2, Shankar seems to just assume (I'm guessing) that $1|V\rangle = |V\rangle$ for all vectors $|V\rangle$ when he does the exercise at the bottom of the page. Is this true, or is it possible to prove $1|V\rangle = |V\rangle$ from the axioms he lists?

 

[CompuChip]

I think that this is what the author means by "do what is natural". Usually, however, 1|v> = |v> is just listed as an axiom ("there is a scalar multiplication r|v> with 1|v> = |v> and 0|v> = |0>").