What are the last two digits of 5^2015

Here are all digits !

Gruß radix !

25 detail detail detail detail

∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity∞infinity

I aM going to think about this

5^2015

$$\\5^1=5\\
5^2=25\\
5^3=125\\
5^4=525\\$$

It seems that if the power is bigger than 1 that the last 2 digiits will always be 25

MAYBE SOMEONE WOULD LIKE TO PROVE THIS  ??

Here's a "proof"

5^2 = 25

5^3  =  25 * 5  =  (20 + 5) (5) = 100 + 25

5^4  = 125 * 5  = (120 + 5) (5)  = (100 + 20 + 5) (5) = 500 + 100 + 25

5^5 = 625 * 5 =  (600 + 20 + 5) (5)  = (500 + 100 + 25)(5) = (500 + 100 + 20 + 5) (5)  = 2500 + 500 + 100 + 25

So it appears that the pattern  for 5^n = 

25 + 100*5^(0) + 100*5^(1) +.....+ 100*5^(n - 4) + 100*5^(n- 3)  for n ≥ 5 

And, in this series, all terms except the first one will end in the digits "00"...so adding 25 to this sum will always result in a number ending in "25"..........

Thanks CPhill