Find the last two digits of $1717^{1717}$ 
$1717^{1717} \mod 100$

Since $\phi(100) = 40$ , we can transform this into:

$17^{37}101^{37} \mod 100 = 17^{37} \mod 100$

How do I proceed further?

Like What are the last two digits of $77^{17}$?,
using Carmichael Function, $\lambda(100)=20\implies**17^{**17}\equiv17^{17}\pmod{100}$
Now $17^2=290-1, 17^{17}=17(290-1)^8$
Again, $\displaystyle(290-1)^8=(1-290)^8\equiv1-\binom81290\pmod{100}\equiv1-90\cdot8\equiv-19$
Now $-19\cdot17=-323\equiv77\pmod{100}$
See also: 
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