Prove a polynomial in Fq is a permutation polynomial of Fqn with a necessary and sufficient condition P.S This is the best Math Expression I can edit. I am real shameful, where can I find the introduction of typing in this webset? thank you 

Exercise7.13 Let\[f\left( x \right) = \sum\limits_{i = 0}^{n - 1} {{c_i}{x^{{q^i}}} \in {\Bbb{F}_q}[x]} \]

Prove the f is a permutation polynomial of \[{\Bbb{F}_{{q^n}}}\] if and only if \[\gcd \left( {\sum\limits_{i = 0}^{n - 1} {{c_i}{x^i},{x^n} - 1} } \right) = 1\] 
I think there is a Theorem , in the front of the Exercise ,that is helpful to solve this exercise. Theorem 7.10 : Let \[r \in N\]with \[\gcd \left( {r,q - 1} \right) = 1\] and let  s  be a positive divisor of \[q-1.\] Let \[g \in {\Bbb{F}_q}\left[ x \right]\] be such that\[g\left( {{x^s}} \right)\]has no nonzero root in $\Bbb{F}_q$. Then \[f\left( x \right)={{x}^{r}}{{\left( g\left( {{x}^{s}} \right) \right)}^{{\left( q-1 \right)}/{s}\;}}\] is a permutation polynomial.

(And I do not know how to use the condtion of the exercise.)
 A: A solution comes from the theory of linearized polynomials. See e.g. Lidl & Niederreiter. I describe the main steps.
The first thing we observe that the polynomial $f(x)$ is $q$-linear. If we denote the algebraic closure of $\Bbb{F}_q$ by $\overline{F}$, then the polynomial mapping $f:\overline{F}\to\overline{F}$ is linear over the field $\Bbb{F}_q$. In other words, we have
$$
f(\alpha x+\beta y)=\alpha f(x)+\beta f(y)
$$
for all $\alpha,\beta\in\Bbb{F}_q$ and all $x,y\in\overline{F}$. This is because $f$ is an $\Bbb{F}_q$-linear combination of the powers of the Frobenius automorphism $x\mapsto x^q$ that fixes elements of $\Bbb{F}_q$ elementwise. Clearly the restriction of this polynomial mapping to $\Bbb{F}_{q^n}$ maps this extension field to itself.
So this polynomial mapping is a linear transformation of an $n$-dimensional space over $\Bbb{F}_q$ to itself. From elementary linear algebra we recall that such a mapping is bijective (i.e. a permutation) if and only if it is non-singular. Or equivalently that its kernel is trivial. 
The elements of $\Bbb{F}_{q^n}$ are precisely the roots of another $q$-linearized polynomial $g(x)=x^{q^n}-x$. The elements of the kernel of $f\mid_{\Bbb{F}_{q^n}}$ are thus exactly the zeros of
$$h(x)=\gcd(x^{q^n}-x,f(x)).\qquad(*)$$
Thus the said kernel is trivial if and only if $h(x)=x$.
To find a necessary and sufficient condition for $(*)$ we use the theory of conventional and linearized associates of $q$-linearized polynomials. If we are given a (conventional) polynomial $p(x)=\sum_{0\le i\le m}p_ix^i\in\Bbb{F}_q[x]$, then we call the $q$-linearized polynomial
$$
\tilde{p}(x)=\sum_{i=0}^mp_ix^{q^i}
$$
the linearized associate of $p(x)$. Similarly we call $p(x)$ the conventional associate of $\tilde{p}(x)$. A general result (see Lidl&Niederreiter) that you may be expected to use here is that passing to the linearized associates commutes with forming the greates common divisor. Or
$$
\gcd(p_1(x),p_2(x))=p_3(x)\Longleftrightarrow \gcd(\tilde{p_1}(x),\tilde{p_2}(x))=\tilde{p_3}(x).
$$
This result settles your question immediately. After all, the given $f(x)$ is the linearized associate of $\sum_ic_ix^i$, and $g(x)$ is the linearized associate of
$x^n-1$. Thus $(*)$ gives
$$\gcd(f,g)=x=\tilde{1}\Longleftrightarrow \gcd(x^n-1,\sum_ic_ix^i)=1.
$$
