As the title suggests, we need to
Show that $p(x)=x^6+\cdots+x^2+x+1$ is irreducible over $\Bbb{Z_{17}}$
We can immediately answer that it is indeed irreducible since it is the cyclotomic polynomial $\Phi_7(x)$, thus it is ireducible over any field $\Bbb{F_p}$, with $p\mod7$ being a generator of the multiplicative group of $\Bbb{Z_7}$.
Alternatively, we can notice that $$x^7-1=(x-1)(x^6+\cdots+x^2+x+1)=(x-1)p(x)$$ and if $r$ is a zero of $p(x)$, it follows that it must have an order of either $1$ or $7$. Since the unit element $1$ is the only one with order one, and we can easily check that it is not a zero, any root-if it exists-must be of order $7$.But since the order of $\Bbb{Z_{17}}$ is $16$ and $7$ does not divide $16$, $p(x)$ has no roots in $\Bbb{Z_{17}}$.
We can also verify that it has no quadratic or cubic factors (we have already shown that no linear factors exist), so it is not irreducible over $\Bbb{Z_{17}}$. (This will involve considering an algebraic extension, of order $2$ and $3$ with $17^2-1$ and $17^3-1$ elements respectively and checking that by Lagrange's Theorem, the order of the element would have to divide $17^2-1=288$ and $17^3-1=4912$, but $7$ does not divide either of the two, thus it has no quadratic or cubic factors).
What I am looking for is something more trivial, if such a solution exists.
Show that $p(x)$ is a factor of $x^{17^6}-x$ and relatively prime to $x^{17^3}-x$ and $x^{17^2}-x$.
To do so, note that $p(x)(x-1)=x^{7}-1$. So $x^{17^2}-x=(x^7-1)q(x)+x^{2}-x$ for some $q(x)$. Similarly, note that $x^{17^3}-x=(x^7-1)r(x)+x^6-x$.
So you just need to show that $p(x)$ is relatively prim to $x^2-x$ and $x^6-x$. This is easy to do.
Finally, $x^{17^6}-x$ is actually divisible by $x^7-1$ since $17^6-1$ is divisible by $7$.