Proof of $\operatorname{Br}(K) = H^2(G,K^*)$

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The proofs I know of the fact $\operatorname{Br}(K) = H^2(G,K^*)$ ($G= \operatorname{Gal}(K^s/K))$ involve non-abelian group cohomology of $H^1(G,PGL_n(K))$. Are there any nice conceptual proofs which don't non-abelian cohomology in anyway?

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Yes. You can use crossed product algebras to do this. See the book by Pierce on Associative Algebras, for example.