Prove that there are no primes in the following infinite sequence of numbers: $1001, 1001001, 1001001001, 1001001001001, ...$ 
Prove that there are no primes in the following infinite sequence of
  numbers: $$1001, 1001001, 1001001001, 1001001001001, ...$$

The sequence can be expressed as follows: $$ f: \mathbb{N}-\{0\} \to \mathbb{R} \\ k \to \frac{1-1000^{k+1}}{1-1000}$$
I thought for a moment to use the recurrence relation $a_0=1$ and $a_k = 1000 a_{k-1} + 1$, but it looks like it led nowhere.
Is it possible to use the explicit function? I don't know how to continue. Will someone be able to give me advice? Please, I ask you not to give me the answer. I would like to do exercise myself.
 A: Hint : Look at the polynomial $p_n(x)= 1 + x +\dots+ x^n$ for $n= 1, 2,\cdots$ And notice the numbers in the sequence are precisely $p_n(10^3)$. Now by the formula you wrote, we have:
$(10^3 -1) p_n(10^3)= (10^{3(n+1)}-1)$. 
Now I think a divisibility argument should work. Let me know if anything is unclear.
A: This is the complement answer of Aranya Lahiri.
We have $(10^3 -1) f(k)= (10^{3(k+1)}-1) =  (10^{k+1}-1)(10^{2k+2}+10^{k+1}+1) $.
If $f(k)$ is prime, so $f(k) \mid (10^{k+1}-1) $ or $f(k) \mid (10^{2k+2}+10^{k+1}+1)$.
Each element $a_k$ of the sequence can be written as $a_k = 1000a_{k-1} + 1$ for a certain $k$, where $a_0=0$. 
But $a_k \equiv 1 \pmod 4$ and $(10^{k+1}-1) \equiv -1 \equiv 3 \pmod 4$, thus $f(k) \nmid (10^{k+1}-1) $.
If $f(k) \mid (10^{2k+2}+10^{k+1}+1)$ $\implies$ there exists $t \in \mathbb{Z}$ such that $\frac{1-10^{3(k+1)}}{1-10^3}t = 10^{2k+2}+10^{k+1}+1$. But $t$ is simply a multiple of $\frac{1-10^3}{1-10^{k+1}}$ as we have already seen. The only case where $\frac{1-10^3}{1-10^{k+1}}$ is an integer is when $f(k=2)$, but this number is also divible by $3$. Consequently, $f(k) \nmid (10^{2k+2}+10^{k+1}+1)$. (Contradiction)
Are there any corrections to this answer?
A: Let the sequence be $a_n$. Obviously,  $a_1=7\times 11 \times 13$ and $a_2= 3\times 333667$ are not primes. Now let’s consider $a_n$ for $n\geq 3$.
$$\displaystyle \begin{aligned}a_n & =1+10^3+10^6+\ldots+10^{3n} \\& =\frac{\left(10^3\right)^{n+1}-1}{10^3-1} \\&=\frac{\left(10^{n+1}\right)^3-1^3}{999}\\&=\frac{\left(10^{n+1}-1\right)\left(10^{2 (n+1)}+10^{n+1}+1\right)}{999} \end{aligned}\tag*{} $$
For any $n\geq 3$, both $10^{n+1}-1$ and $10^{2 (n+1)}+10^{n+1}+1$ are greater than $999$. So $a_n$ can be factorised as a product of two integers greater than $1$ and hence $a_n$’s are never prime numbers.
