<s>[INST] Is it true that $76^n=76\pmod{100}$ for all $n>0$? Note
$76^2=5776,$
$76^3=438976,$
$76^4=33362176,$
$76^5=2535525376,$
$$\cdots$$
Hence I guess that 
$$76^n = 76 \pmod{100} $$
for any natural number $n$.
Is it correct?
And are there other numbers that have this property?
 [/INST] It follows by induction:
If $a^2 \equiv a \mod n$ then $a^k \equiv a \mod n$ for all natural numbers $k$.
Pf:  Base case:  $a^1 \equiv a \mod n$.
Inductive step: If $a^n \equiv a \mod n$ then $a^{n+1} =a^n*a\equiv a*a=a^2 \equiv a \mod n$.
Even if $a^2 \not \equiv a \mod n$ but if you have instead $a^{m+1}\equiv a \mod n$ then $a^{km +1}\equiv a \mod n$ for all $k$.
example $51^2=2601 \equiv 1 \mod 100$ and $51^3 =132651\equiv 51 \mod 100$ so $51^{1 + 2k} \equiv 51 \mod 100$.
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To find out if there are others we need $(10a + b)^2 = 100a + 20ab + b^2$ to end with $10a + b$.  
Only $0, 1, 5,6$ when squared will end with itself.
$(10a + 0)^2 = 100a^2$ will end with $00$ so $0$ is one such number.
$(10a + 1)^2 = 100a^2 + 20a + 1$ so we need $20a$ to end with $a$ so $a = 0$ and $1$ is another such number.
$(10a + 5)^2 = 100a^2 + 100a + 25$ so we need $a=2$ and so $25$ is another such number.
Finally $(10a + 6)^2 = 100a^2 + 120a + 36$ so we need $2a + 3$ to end with $a$.
If $a = 0,1,2,3,4,5,6,7,8,9$ then $2a + 3 = 3,5,7,9,11,13,15,17,19,21$.  Only $2*7 + 3 = 17$ will end with $a$.  So $76$ is the last such number.
$0,1,25$ and $76$ are the only such numbers.
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