Prove if $L = K(α_1, . . . , α_r)$ and each $\alpha_i$ is separable over $K$, then $L/K$ is separable 
Let $L/K$ be a finite extension, $[L:K] = n$.
Prove the following are equivalent:

*

*$L/K$ is separable


*$L=K(\alpha_1,...,\alpha_n)$ and every $\alpha_i$ is separable over $K$.

I understand the part where $(1\ \implies 2)$, can anyone please explain to me the other direction $(2 \Longleftarrow 1)$?
 A: Claim 1: Let $\alpha$ be algebraic over $K$. $K(\alpha)/K$ is separable if, and only if, $\alpha$ is separable over $K$.
Proof: (idea) As $\alpha$ is algebraic over $K$ we may take $p_{\alpha} (x)$ irreducible polynomial such that $p_{\alpha} (\alpha) =0$. Notice that $K(\alpha) / K$ has degree $n = \partial p_{\alpha}$ and $[K(\alpha):K]_s$ is the number of distinct roots of $p_u(x)$ (look at the closure if necessary). Those numbers are the same if, and only if, $p_u(x)$ is separable. If $p_{\alpha}(x)$ is irreducible then $p'_{\alpha}(x) \neq 0$ then is separable. 
Claim 2: Let $\alpha_1, \ldots , \alpha_n$ be algebraic over $K$. 
$K(\alpha_1, \ldots , \alpha_n)/K$
 is separable if, and only if, $\alpha_1, \ldots , \alpha_n$ are separable over $K$.
Proof: Assuming $\alpha_1, \ldots , \alpha_n$ are separable over $K$ we will use induction over $n$.


*

*$n=1$ See Claim 1;

*Considering the assertion true to $n-1$ and the tower 
$$K \subseteq K(\alpha_1, \ldots , \alpha_{n-1},\alpha_n) \subseteq K(\alpha_1, \ldots , \alpha_{n-1})(\alpha_n)$$
it follows that $K(\alpha_1, \ldots , \alpha_{n-1})/K$ is separable by induction hypothesis. Now $\alpha_n$ is separable over $K$ hence is it is also separable over $K(\alpha_1, \ldots , \alpha_{n-1})$ (look at the minimal polynomial over the larger field). By Claim 1 we see that $$K(\alpha_1, \ldots , \alpha_{n-1}, \alpha_n)/K(\alpha_1, \ldots , \alpha_{n-1})$$ 
is separable, then $K(\alpha_1, \ldots , \alpha_n)/K$ is separable. 
