Last digits, numbers Can anyone please help me?
1) Find the last digit of $7^{12345}$ 
2) Find the last 2 digits of $3^{3^{2014}}$.
Attempt: 1)
By just setting the powers of $7$ we have $7^1 = 7$, $7^2=49$, $7^3=343$, $7^4 = 2401$, $7^5 = 16807$, $7^6 = 117649$, $\dots$ 
After the power of $4$, the last digits will repeat. Then by noticing the pattern the digits will end in $7,9,3$ and $1$.
Then we can divide the exponent $(12345)$ by $4$ since this is the cycle that makes it repeat. Then $12345 : 4$ has remainder $1$. So $7^1 = 7$ is the unit digit to $7^{12345}$. 
So the last digit is $7$. 
I know how to do it like this, the problem does not state how to find the last digit, but I know it has something do do with Euler's theorem.
for part $2$) I don't know how to start. Can anyone please help me? 
Thank you for the help.

Let $$3^{2014}-1=2x$$
Your number is : $3^{2x+1}=3\cdot3^{2x}=3(10-1)^{x}=3(-1+10)^{x}$
Using binomial theorem and neglecting powers of $10$ greater than $2$ as we want only last $2$ digits.
$$3(-1^{x}+x(-1)^{x-1}10)$$
Writing $3$ as $4-1$ in first expression will lead you to believe that $2x$ is divisible by $4$. Hence, $x$ is even.
$$3(1-10x)=3(1-5\cdot2x)$$
I hope you can carry on from here. You will need last two digits of $3^{2014}$ which you can find by writing it as $(10-1)^{1007}$
