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.

First of all, finding last $n$ digits in Base $10$ essentially finding modulo $\displaystyle10^n$
Using Carmichael function, $\displaystyle3^{\lambda(100)}\equiv1\pmod{100}$ as $(3,100)=1$
$$\implies3^{3^{2014}}\equiv3^{3^{2014}\pmod{\lambda(100)}}\pmod{100}$$
Now $\displaystyle\lambda(100)=20,$ so we need to find $\displaystyle3^{2014}\pmod{20}$
and by observation $\displaystyle3^4=81\equiv1\pmod{20}$ or using Carmichael function $\displaystyle\lambda(20)=4\implies3^4\equiv1\pmod{20}$
and as $\displaystyle2014\equiv2\pmod4\implies3^{2014}\equiv3^2\pmod{20}\equiv9$
$$\implies3^{3^{2014}}\equiv3^9\pmod{100}$$
Now $\displaystyle3^9=3\cdot3^8=3\cdot(3^2)^4=3(10-1)^4=3(1-10)^4=3\left[1-\binom4110+\binom4210^2-\binom4310^3+10^4\right]$
$\displaystyle\equiv3\left(1-40\right)\pmod{100}\equiv-117\equiv83$
