Find the last 2 digits of 7^2014?

I am still getting used to these questions and I still like them. :)

I'd do it the same as CPhill but I would start with a smaller number.

$$\\7^1=7\\ 7^2=49\\ 7^3=343\\ 7^4=2401\\ $that is a really helpful one$\\ $(Any number ending is 01) raised to any positive integer n must also end in 01 because only the last 2 digits will affect the outcome and 1^n=1 $ \\\\ (7^4)^{n} $ must end in 01 for any positive integer n$\\\\ 7^{2014}=7^{2012}*7^2\\ 7^{2014}=7^{4*503}*7^2\\ 7^{2014}=(7^4)^{503}*49\\ 7^{2014}= ....01 *49\\ 7^{2014}= ....49\\$$

so the last 2 digits will be 49    Just like CPhill said 

Notice that

7^20  ends in 01

So 

(7^20)^5  = 7^100 also ends in 01

And

7^2000 = (7^100)^20   will also end in 01  

And 7^14   ends in 49

So   7^2014 =   7^2000 x 7^14  =      .........01  x  ........49  

 Will  also end in 49