How can I find the units digit of $2^{11213}-1$ Question: in 1963 mathematicians at the University of llinois used a computer to show that $2^{11213} -1$ is prime. this number has 3376 digits in base 10. what is the units digit?
I haven't learned any modular arithmetic or Euler's function. I've been given this as a bonus problem in my precalculus class. I was thinking that I could look for some repeating patters for the last digit with powers of 2. 
$2^1=2$
$2^2=4$
$2^3=8$
$2^4=6$
$2^5=2$
So it repeats every 4 powers. However I do not know what to do from here or if i'm on the correct track
 A: You're on the correct track.  Likewise, you could say $2^{11213}=2\times(2^4)^{2803},$ and $2^4$ ends with $6,$ and powers of numbers ending with $6$ always end with $6$, so $2\times(2^4)^{2803}$ ends with $2$, and that prime ends with $1$.
A: 
So it repeats every 4 powers. 

Yes!!!!!!!

However I do not know what to do from here

Oh no!!!!

or if i'm on the correct track

You are SOO much on the right track that if you don't get up and jump under that train, I'll pick you up and throw you under!  And .... oh, wait.... that's not a very encouraging metaphor at all!
....
Okay, you are soooo close.
If it repeats every $4$ powers you have the last digit of $2^k$ is the same last digit of $2^{k+ 4}$ is the last digit of $2^{k+8}$ is the same last digit of ....  $2^{k + 4m}$.... for any $m$.  
And so because $11213 = 4*2803 + 1$ then $2^{11213} = 2^{1+4*2803}$ will have the same last digit as $2^{1+4*2802}$ has the same last digit as $2^{1+4*2801}$ has ..... the same last digit as $2^{1+4*3}$ has the same last digit as $2^{1+4*2}$ has the same last digit as $2^{1 + 4}$ has the same last digit as  $2^1$.
So the last digit of $2^{11213}$ is $2$ and the last digit of $2^{11213} -1$ is $1$.
A: Note that the exponent is at form $4k+1$. So what is the unit of power?
From the pattern you found, $2$, right?
So, the unit of the number is $2-1=1$.
A: It is easy to prove that $$2^{4k+1}\equiv2\pmod{10}\\2^{4k+2}\equiv4\pmod{10}\\2^{4k+3}\equiv8\pmod{10}\\\\2^{4k}\equiv6\pmod{10}$$
Then, since $11213=4k+1$ the power $2^{11213}$ has last digit $2$ and  $2^{11213}-1$ has $\color{red}1$ as unit digit.
