Can $7$ be the smallest prime factor of a repunit? Repunits are numbers whose digits are all $1$. In general, finding the full prime factorization of a repunit is nontrivial.
Sequence A067063 in the OEIS gives the smallest prime factor of repunits. There are no $7$'s (just the number $7$, not as a digit in a different prime) that I can see in the sequence.

$7$ does divide many repunits, but always does when $3$ is also a factor, so $7$ never shows up as the smallest prime factor. The table of the first $508$ smallest prime factors of repunits shows that $7$ is not the smallest prime factor of any of those repunits.
My question is can $7$ ever be the smallest prime factor of a repunit?
I tried to find a repunit with $7$ as the smallest factor but searched very far and $3$ is always a factor whenever $7$ is.
I tried to prove that it can't be but I have no idea how.
 A: 
Can $7$ be the smallest prime factor of a repunit?

No, it cannot. 
The main idea is as follows: If the $n^\text{th}$ repunit $R(n)$ is divisible by $7$, then $n$ is divisible by $6$, which means $n$ is divisible by $3$, which means $R(n)$ is divisible by $3$, so $7$ cannot be the smallest prime factor. 
The full argument is fleshed out below. 

Let $R(n)$ be the $n^\text{th}$ repunit, e.g. $R(4) = 1111$. 
One interesting property about repunits is that if an integer $m$ divides $R(n)$, then $m$ divides $R(2n)$, and $m$ divides $R(3n)$, and so on. This is because, for any positive integer $n$,  we have the factorization $$R(2n) = \left(10^{n}+1\right)R(n)$$
and, more generally, for any positive integers $k$ and $n$, we can factor as follows:
$$R(kn) = \left(\displaystyle\sum\limits_{j=0}^{k-1} 10^{jn}\right)R(n)$$

What this means, for your problem, is if a prime $p$ divides a repunit $R(n)$, then it will also divide $R(kn)$ for all $k$. 
This tells us that:


*

*Every other repunit is divisible by $11$.  $$11 \,|\, R(2) \implies\,\Big(\,\,\,n \equiv 0 \pmod{2} \iff 11 \,|\, R(n)  \,\,\,\Big)$$ Here, the "$\iff$" bidirectionality follows from the fact that $11 \not| \,\,R(1)$. 

*Every third repunit is divisible by $3$. $$3 \,|\, R(3) \implies \,\Big(\,\,\,n \equiv 0 \pmod{3} \iff 3 \, |\, R(n) \,\,\,\Big)$$ Here, the "$\iff$" bidirectionality follows from the fact that $3  \not| \,\,R(1)$ and $3  \not| \,\,R(2)$. 

*Every third repunit is divisible by $37$. $$37 \,|\, R(3) \implies \,\Big(\,\,\,n \equiv 0 \pmod{3} \iff 37 \, |\, R(n) \,\,\,\Big)$$ Here, the "$\iff$" bidirectionality follows from the fact that $37  \not| \,\,R(1)$ and $37  \not| \,\,R(2)$.

*Every sixth repunit is divisible by $13$.$$13 \,|\, R(6) \implies \,\Big(\,\,\,n \equiv 0 \pmod{6} \iff 13 \, |\, R(n) \,\,\,\Big)$$ Here, the fact that each of $1, 11, 111, 1111, 11111$ are not divisible by $13$ implies the "$\iff$" bidirectionality. 

*Every sixth repunit is divisible by $7$.$$7 \,|\, R(6) \implies \,\Big(\,\,\,n \equiv 0 \pmod{6} \iff 7 \, |\, R(n) \,\,\,\Big)$$ Here, the fact that each of $1, 11, 111, 1111, 11111$ are not divisible by $7$ implies the "$\iff$" bidirectionality. 

This last fact implies that $7$ can never be the smallest prime factor of a repunit, because whenever $n \equiv 0 \pmod{6}$, it's also true that $n \equiv 0 \pmod{3}$, so whenever $7$ is a factor of a repunit, $3$ will be as well. 
By a similar argument, $13$ and $37$ also cannot be the  smallest prime factor of a repunit. 
A: $7|1001 = 7*11*13$
So $7|1001*111 = 111,111$ so if $7|\underbrace{1111.....1}_n$ then
$7|\underbrace{11111.....1}_n - 111111$
$ = \underbrace{1111111....111}_{n-6}000000$
and as $10$ and $7$ are relatively prime. $7|\underbrace{1111111....111}_{n-6}$
So by induction if $n \equiv a \pmod 6$ then $7|\underbrace{1111...1}_{a}$.
So we just need to show when does $7|1,11,111,1111,11111,111111$
And that is only $7|111111 = 111*1001 = (3*37)*(7*11*13)$.[*]
So $7|\underbrace{1111......1}_n$ if and only if $6|n$ in which case $3,11,13,37$ will also divide the repunit.
Thanks to Zubin Mukerjee for the hint.
.....
[*].   $7|\underbrace{111...1}_{a < 6}\iff 7|\underbrace{111...1}_{6-a}$ but $7\not \mid 1(a=1)$ and $7\not \mid 11 (a=2)$ and so $7\not \mid 11111 (6-1)$ and $7\not\mid 1111 (6-2)$ while $7\not \mid 111 =3*37$ 
