Proof, that polynomial $\prod_{i=1}^{n} (X - a_i) + 1 \in K[X]$ doesn't have roots We have $n$ different elements $(a_1,...,a_n)$ that are all the elements of $K$ and  $\in$  finite field $K$.
I want to prove, that  $\prod_{i=1}^{n} (X - a_i) + 1 \in K[X]$ doesn't have roots
I know, that if $a_i$ is a root of polynomial $p \in K[X]$ , then exists $f \in K[X]$ such that $p = (x - a_i)f$
How can we use this fact in order to prove the statement?
 A: If $\;\{a_1,...,a_n\}\;$ are not all the elements of $\;K\;$ the claim still is false even if $\;K\;$ is a finite field, for example:
$$K=\Bbb F_3\;,\;\;a_1=0,\,a_2=1\implies g(x)=x(x-1)+1=x^2-x+1$$
and still $\;g(2)=0\;$ ...
If $\;a_1,...,a_n\;$ are all the elements of the finite field $\;K\;$ then the claim is true, since
$$g(x)=\prod_{k=1}^n(x-a_k)+1\implies g(a_j)=1\neq0\;,\;\;\forall\,j=1,2,...,n$$
A: Let $k$ be a finite field, and set $p(x) = 1 + \prod_{\alpha\in k}(x - \alpha)\in k[x]$. Then a root of $p$ is simply an element $\beta$ (say, $\beta\in\bar{k}$ some algebraic closure of $k$) such that $p(\beta) = 0$. If $\beta\in k$, then
\begin{align*}
p(\beta) &=1 + \prod_{\alpha\in k}(\beta - \alpha)\\
&= 1 + (\beta - \beta)\cdot\prod_{\alpha\in k\setminus\{\beta\}}(\beta - \alpha)\\
&= 1 + 0\cdot\prod_{\alpha\in k\setminus\{\beta\}}(\beta - \alpha)\\
&= 1.
\end{align*}
Since $1\neq 0$, $p$ has no roots in $k$. Of course, any polynomial with coefficients in a field has roots in the algebraic closure of that field, so $p$ has roots in this sense.
If you really want to use your fact, you can proceed by contradiction: suppose $\beta\in k$ is a root of $p$. Then $p(x) = (x - \beta)q(x)$ for some $q\in k[x]$. But then
\begin{align*}
1 &= p(x) - \prod_{\alpha\in k}(x - \alpha)\\
&= (x - \beta)q(x) - (x - \beta)\prod_{\alpha\in k\setminus\{\beta\}}(x - \alpha)\\
&= (x - \beta)\left(q(x) - \prod_{\alpha\in k\setminus\{\beta\}}(x - \alpha)\right).
\end{align*}
This is a contradiction, as $r(x) = (x - \beta)\left(q(x) - \prod_{\alpha\in k\setminus\{\beta\}}(x - \alpha)\right)$ is $0$ when $x = \beta$, but $1\neq 0$.
