Prove: for any $n$, $1+10^4+...+10^{4n}$ is a composite number. I saw the first three terms: $1$, $10000$, $100000000$. I thought about proving by induction, so I did:
Base case: At $n>0$, $10001$ is composite. Note that at $n=0$, we get $1$ which is not composite. And at $n<0$, we get decimals which are against composite number definition. So, $n$ cannot be $\le 0$.
Hypothesis: Let it be true for $k$, such that $m = 1+10^4+...+10^{4k}$ is a composite number.
Proof: Add $10^{4(k+1)}$ to $m$. So, m becomes $1+10^4+...+10^{4k}+10^{4k+4}$ which is a composite number + composite number. However, as I can trivially infer from $9+4=13$, a composite number + composite number is not always another composite.
Question: How to solve this question? Hints for getting started?
UPDATE:
So, as @Michael told me, this is a geometric series. So, using $\frac{a(1-r^m)}{1-r}$, so letting $a=1, r = 10^4$, I got: $$\frac{1*(1-(10^{4})^{n+1})}{1-10^4} \implies \frac{10^{4n+4}-1}{9999}$$ How to proceed further?

We write the number as
$$\frac{10^{4m}-1}{10^4-1},$$
where $m \geqslant 2$.
Case 1: $m$ even. Write $m = 2r$, then our number is
$$\frac{10^{8r}-1}{10^4-1} = \frac{10^{4r}-1}{10^4-1}\cdot (10^{4r}+1).$$
If $r = 1$, then it is easy to check that $10^4+1$ is composite, and for $r > 1$ we note that $\frac{10^{4r}-1}{10^4-1} > 1$, which shows our number is composite.
Case 2: $m$ odd. Write
$$\frac{10^{4m}-1}{10^4-1} = \frac{10^{2m}+1}{101}\cdot \frac{10^m+1}{11}\cdot\frac{10^m-1}{9}.$$
Since $m$ is odd, we have $10^{2m} \equiv (-1)^m \equiv -1\pmod{101}$, and $10^m \equiv (-1)^m \equiv -1\pmod {11}$, so all factors are integers. Since $m > 1$, all three factors are $> 1$.
