As part of a course in Algebraic Geometry, I am trying to prove a corollary of a given lemma:

**Lemma:** If $K$ is a field and $\alpha_1,\dots, \alpha_m$ are algebraically independent in $K$ and $\beta \in K$ is algebraically dependent of $\alpha_1, \dots \alpha_m$ but not on $\alpha_1, \dots \alpha_{m-1}$, then $\alpha_m$ is algebraically dependent on $\alpha_1,\dots, \alpha_{m-1}, \beta$.

**Corollary:** If $K$ is a field and $\alpha_1, \dots,\alpha_n$ are algebraically independent, and $\beta_1, \dots, \beta_n$ are algebraically independent (both in $K$) and $\beta_i \in K(\alpha_1,\dots, \alpha_n)$ then $m \leq n$.

My notes say "Use the (Exchange) lemma to replace the $\alpha_i$ by $\beta_i$ one at a time".

I've tried applying the lemma to say:

Re-label if necessary so that $\beta_1$ is algebraically dependent on $\alpha_1,\dots,\alpha_r$ but not on $\alpha_1,\dots, \alpha_{r-1}$ and replace $\alpha_r$ with $\beta_1$. Relabel again so that we now have $\alpha_1,\dots,\alpha_{n-1},\beta_1$, algebraically independent since in any polynomial containing $\beta_1$ and the $\alpha_i$ we can sub in the expression for $\beta_1$ as an element of $K(\alpha_1,\dots, \alpha_n)$ to get an algebraic dependence on the $\alpha_i$. By the lemma the removed $\alpha_n$ is algebraically dependent on $\alpha_1,\dots,\alpha_{n-1},\beta_1$ I'm not sure where to go from here or even if this is the right line of approach, any help would be much appreciated.