If $K \leq L$ a finite extension then it is algebraic. I am looking at the proof of

If $K \leq L$ a finite extension then it is algebraic.

The proof is the following:
Let $[L:K]=n<\infty$. Let $a \in L$.
We will show that $\exists$ a non-zero $f(x) \in K[x]$ with $f(a)=0$.
We take $1, a, a^2, \dots , a^n \in L$, that are $(n+1)$ elements of $L$.
So, it is $K-$linear dependent, so there are $k_0, k_1, \dots , k_n \in K$ not all zero, with $k_0 \cdot 1 +k_1 \cdot a+k_2 \cdot a^2 + \dots + k_n a^n=0$.
Then $f(x)=k_0+k_1x+ \dots + k_nx^n \in K[x]$ is non-zero and $f(a)=0$.
Therefore, $a$ is algebraic over $K$.
$$$$
Could you explain me the following part??

We take $1, a, a^2, \dots , a^n \in L$, that are $(n+1)$ elements of $L$.
So, it is $K-$linear dependent, so there are $k_0, k_1, \dots , k_n \in K$ not all zero, with $k_0 \cdot 1 +k_1 \cdot a+k_2 \cdot a^2 + \dots + k_n a^n=0$.

 A: The operations $$\begin{align}L \times L &\to L\\(u,v) &\mapsto u + v\end{align}$$ and $$\begin{align}K \times L &\to L\\(\lambda,u) &\mapsto \lambda u\end{align}$$
exist naturally in $L$. The properties that define a  vector field hold (a simple exercise). So we may talk about a $K$-vector space. 
Notice that if $a \in L$ we have that $[K[a]K;] \leq n < \infty$ is a subspace of $L$. Then if $[K[a]; K] = m$ we have that 
$$\{1,a,\ldots,a^{m-1} \}$$
is a basis* for $K[a]$, so $1,a,\ldots, a^m$ are, in fact linearly dependent, then there exist $k_0,k_1,\ldots, k_m \in K$ not all zero that $$k_0 1+k_1a+\ldots+k_ma^m = 0$$
Take the polynomial $f(X) = k_0 + k_1X + \ldots + k_mX^m$.     
(*) It comes from this if $f(X) \in K[X]$ and $p_{a,K}(X) \in K[X]$ is irreducible then there exist $q(X), r(X) \in K[X]$ such that $$f(X) = q(X)p(X) + r(X)$$
where $r(X) = 0$ or $\partial r(X) < \partial p(X)$. 
So $$f(a) = q(a)p(a) + r(a) = r(a) = k_o + k_1a + \ldots+ k_{m-1}a^{m-1}$$
that is, any polymonial can be written as the latter form, then (after checking unicity of writing) we may see that $$K[a] = \{k_a + k_1a+\ldots+k_{m-1}a^{m-1}; k_i \in K\}$$
