Proof of an Elementary Result in Linear Algebra Is the Following Proof Correct?
Theorem. If the vectors $\alpha_1,\alpha_2,...,\alpha_n$ constitute a linearly independent list in the Vector space $V$ where as the list of vectors $\alpha_1,\alpha_2,...,\alpha_n,\beta$ is linearly dependent in $V$ then $\beta$ can be uniquely expressed as a linear combination of $\alpha_1,\alpha_2,...,\alpha_n$
Proof. Let $\beta = \alpha_{n+1}$. Now since the list $\alpha_1,\alpha_2,...,\alpha_n,\alpha_{n+1}$ is linearly dependent it follows that there exists some $j\in I = \{1,2,3,...,n,n+1\}$ such that $\alpha_j\in\operatorname{span}(\alpha_1,\alpha_2,...,\alpha_{j-1})$, evidently this $j = n+1$ since assuming that $j\in I\backslash\{n+1\}$ contradicts the fact that the list $\alpha_1,\alpha_2,...,\alpha_n$ is linearly independent in $V$. 
We therefore conclude that $\alpha_{n+1} = \beta\in\operatorname{span}(\alpha_1,\alpha_2,...,\alpha_n)$, further more the unique representation of $\beta$ as a linear combination of $\alpha_1,\alpha_2,...,\alpha_n$ follows from the fact that $\alpha_1,\alpha_2,...,\alpha_n$ is linearly independent in $V$.
$\blacksquare$

Yes, it is correct. However, you can further explain the assumption $j\in I\backslash \{n+1\}$ and why you don't lose generality.
