This is a corollary of the soundness theorem, which states that for a set of formulas $\Phi$ (of propositional logic) and a formula $\alpha$ : $$\Phi\vdash\alpha\Longrightarrow\Phi\vDash\alpha$$ What I'm trying to show is the case where $\Phi=\emptyset$, i.e. that if $\alpha$ is a proof then it must be a tautology, but without applying the soundness theorem directly.
A proof for the soundness theorem could go like this: let $\Phi\vdash\phi$, then there is a deduction $\phi_1,\ldots,\phi_n$ from $\Phi$ of $\alpha=\phi_n$. Each $\phi_k$ of the deduction is either a tautology, a member of $\Phi$ or the result of Modus Ponens of two previous $\phi_i$ and $\phi_j$. We can then show by strong induction on $k$ that $\Phi\vDash\phi_k$.
However, I don't quite see how we could apply this scheme to the case $\Phi=\emptyset$. Any idea? Or another way of showing it?