About polynomials. If there are no two polynomials of the same degree in $S$, then $S$ is linearly independent. The problem states as follows.

Let $S$ be a set of non-zero polynomials over a field $ F$. If there
  are no two polynomials of the same degree in $S$, then $S$ is linearly
  independent.

I tried the following.
Let be $f_1,\ldots,f_n$ be polynomials chosen such that if $i<j$ then $\deg f_i < \deg f_j$.

If we put $$c_1f_1+\ldots+c_nf_n=0$$
  then $$c_nf_n=c_1f_1+\ldots+c_{n-1}f_{n-1}.$$
If $c_n\neq 0$, then $c_nf_n\neq 0$ so $c_1f_1+\ldots+c_{n-1}f_{n-1}\neq 0$ but we know that $$\begin{align}\deg(c_nf_n)&=\deg (c_1f_1+\ldots+c_{n-1}f_{n-1})\\&\leq \max \{\deg f_1, \ldots,\deg f_{n-1}\}\\&=\deg f_{n-1}\\&<\deg f_n\end{align}$$ and this is a contradiction. Then we have that $c_n=0$. 
By applying the same process, we found that $c_1=\ldots=c_{n-1}=0$.
So $S$ is linearly independent.

Is my proof correct? Is there any shortest one? 
Thanks for your help.
 A: Your proof is fine, I suppose. A minor reformulation if one wants to avoid the slightly informal "repeat the process" and "$\ldots$" could be to either write this as proof by induction ($c_n\ne 0$ leads to a contradiction; hence $c_n=0$. But then by induction hypotheses also $c_k=0$ for $1\le k\le n-1$). Or assume that not all $c_i$ are $=0$, let $m$ be maximal with $c_m=0$, then write $f_M$ in terms of the lower degree $f_i$ and arrive at the same contradiction.
A: The given proof looks OK to me.
Here's another pretty brief argument:
set
$S_0 = \{ x^m \mid m \in \Bbb Z, m \ge 0 \} = \{1, x, x^2, \ldots \}; \tag{1}$
then $S_0$ is a linearly independent set, since a linear dependence amongst the elements of $S_0$ would entail the existence of an $n \ge 0$, $n \in \Bbb Z$, and $\alpha_i \in F$, $0 \le i \le n$, not all $\alpha_i = 0$, with
$p(x) = \sum_0^n \alpha_i x^i = 0, \tag{2}$
implying $p(x)$ is an identically vanishing polynomial not all of whose coefficients are $0$;   NOT!!! 
Now if there is some non-trivial linear dependence amongst some non-zero $q_1(x), q_2(x), \dots, q_l(x) \in S$,
$q(x) = \sum_1^l \beta_j q_j(x) = 0, \tag{3}$
we may clearly assume that $\beta_j \ne 0$, $1 \le j \le l$,  and we may without loss of generality assume $\deg q_l(x) > \deg q_j(x)$ for $1 \le j < l$; we also have the leading coefficient of $q_l(x)$, call it $\lambda \ne 0$.  But then (3) is a non-trivial linear dependence amongst the $x^i$, since
the leading coefficient of $q(x)$ is
$\beta_l\lambda \ne 0; \tag{4}$
by virtue of the fact that $\deg q_l(x) > \deg q_j(x)$ for all $j$ other than $l$, none of those $q_j(x)$ can contribute to the coefficient of $x^{\deg q_l}$.  But such a linear dependence is prohibited by the above affirmation of the linear independence of the set $S_0$;  thus $S$ itself is a linearly independent set.
