So many 5s but

Using the Carmichael function, we find that $5^8\equiv{1}\pmod{32}$ . Thus $5^{1234566}=5^{8\times{154320+6}}\equiv5^6\equiv9\pmod{2^5}$ . Multiplying through with $5^5$ we find that $5^{1234571}\equiv9\times5^5=28125\pmod{10^5}$

5^{4}=0625 : 5^{5}=3125 :5^{6}=..5625: 5^{7}=..8125: 5^{8}=..0625: 5^{9}=..3125

.................. :5^{1234568}=....0625

5^{1234569}=...3125:5^{1234570}=...5625 : 5^{1234571}=....8125

5^{7}=..78125 : 5^{11}=..28125 : 5^{15}=..78125 : 5^{19}=..28125 ........... :

5^{1234567}=....78125 : 5^{1234571}=...28125

Simply used Euler's Theorem.

To calculate last 5 digits we need to find the remainder by 100000.

So first we cancel $5^{5}$ .

Now we need to find the remainder when $5^{1234566}$ is divided by 32 . By applying Euler's Theorem we obtain remainder = 9.

Since we had cancelled out 3125, so we need to multiply it back. THIS gives the answer $9\times{3125} = 28125$ .

First calculate $5^{1234566}$ $(mod 32)$ . We get it to be 9. Then multiply 3125 to get the desired result.