Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots 
Let $\overline{a_n \ldots a_1a_0}$ be the decimal representation of $65^k$ for some $k \geq 2$. Prove that the polynomial $a_nx^n+\cdots+a_1x+a_0$ has no rational roots.

I thought about using the rational root theorem. We know that $a_0 = 5$ since $65$ ends in a $5$, so any rational root must be of the form $-\dfrac{5}{a}$ or $-\dfrac{1}{a}$ where $a$ is an integer factor of $a_n$. How can we continue?
 A: An incomplete answer, but I hope it may help in getting the final answer itself.
Let $\frac pq$ be a rational root of $f$. Since $f-f(10)$ has $10$ as a root, it follows that  $f-f(10) = (x-10)g(x)$  for some $g$. Now, let $x = \frac pq$:
$$
\left(\frac pq - 10\right) g\left(\frac pq\right) = -f(10) \neq 0
$$
Multiplying by $q^n$ :
$$
\left(p-10q\right)q^{n-1}g\left(\frac pq\right) = -f(10)q^n
$$
From here, we know that $\deg g = n-1$, so that $q^{n-1}g\left(\frac pq\right)$ is an integer. Hence, $p-10q | q^n f(10)$, and by co-primality, $p-10q | f(10) = 65^k$.
We have that $p|5$ and $q | a_n$. So, $p = 1/5$, and $q$ must satisfy $p-10q | 65^k$, where $-9 \leq q \leq -1$ (the polynomial does not have positive roots by Descartes' rule of signs).
Note that if $p=1$, then for the above $q$, not case matches. However, if $p=5$, then $q=-6$ fits the criteria. So all we have to  do is to check that $\frac{-5}{6}$ is not a rational root, then we are done. In fact, this shows that if the leading coefficient $a_n \neq 6$, then we are done.
A: Let $f_n$ be the polynomial obtained from the digits of $65^n$. The other answer shows that if $p/q$ is a root of this polynomial in lowest terms, then $p-10q|65^n$; along with the bounds on $f_n$, this implies that we must have $p=-5$ and either $q=2$ or $q=6$. That is, the only possible rational roots are $-\frac{5}{6}$ and $-\frac{5}{2}$.
Note that $(6x+5)^n$ is also a polynomial which evaluates to $65^n$ at $x=10$. So we can write
$$
f_n(x)=(6x+5)^n+(x-10)R(x)
$$
for some polynomial $R$.
Setting $x=0$ in this expression, we have
$
5=5^n-10R(0)
$,
and so $R(0)=\frac{5^{n-1}-1}{2}$, which is not a multiple of $5$.
Now, suppose $n \geq 3$ and $x$ is either $-\frac{5}{6}$ or $-\frac{5}{2}$. Then $(6x+5)^n$ has $5$-adic valuation at least $3$ (that is, its numerator is a multiple of $5^3$ when it is written in lowest terms). In contrast, the $5$-adic valuation of $R(x)$ is $0$ (because the $5$-adic valuation of $R(0)$ is $0$) and $(x-10)$ is either $-\frac{5}{6}-10=-\frac{65}{6}$ or $-\frac{5}{2}-10=-\frac{25}{2}$, neither of which has $5$-adic valuation greater than $2$. So $(6x+5)^n$ and $(x-10)R(x)$ have unequal $5$-adic valuations, which means that $f_n(x)=(6x+5)^n+(x-10)R(x)$ cannot possibly be $0$.
It only remains to check that $f_2$ has no rational roots, which is an easy calculation.
