Prove that $6^n=6^{n+5} (mod 100)$ How can we prove that $6^n=6^{n+5} (\text{mod}~ 100)$? I tried by writing $6^{n+5}=7776 \cdot 6^n = 76 \cdot 6^n (\text{mod}~ 100)$ but this approach does not lead to the above result.

Well, if you hit a snag that means either... the thing you are trying to prove is false... or it's not actually a snag.
You say $6^{n+5}\equiv 76*6^n \pmod {100}$ does not give the correct result. 
How do you know $76*6^n\not \equiv 6^n \pmod {100}$?
...
$76*6^n \equiv 6^n \pmod {100} \iff$
$75*6^n \equiv 0 \pmod {100} \iff$
$100 |75*6^n \iff$
$4|3*6^n = 3^{n+1}2^n \iff$ (as $2$ is prime)
$4|2^n \iff$
$n \ge 2$.
So your snag is only a snag if $n < 2$ in which case the result isn't true.
....
Alternatively we can use Chinese Remainder Theorem.
If $76*6^n \equiv k \pmod{100}$ then there is a unqiue solution to 
$k \equiv 76*6^n \equiv 6^n \pmod{25}$ and to
$k \equiv 0 \pmod 4$
If $n = 0$ then $6^n \equiv 1 \pmod 4$ and our solution is  $k = 6^n + 25m$ where $25m\equiv m \equiv 3\pmod 4$. (i.e. $m = 3$ and $6^5 = 7776 \equiv 76 = 1 + 75 = 6^0 + 75 \pmod {100}$).
If $n = 1$ then $6^n \equiv 2\pmod 4$ and our solution is $k = 6^n + 25 m $ where $25m \equiv m \equiv  2\pmod 4$ (i.e. $m = 2$ and $6^6= 46656 \equiv 56\equiv  6^1 + 50 \pmod {100}$).
If $n \ge 2$ then $6^n \equiv 0 \pmod 4$ and ore solution is $k =6^n + 25m$ where $25m \equiv m \equiv 0 \pmod 4$ and $6^{n+5}=7776*6^n \equiv 76*6^n = 75*6^n + 6^n \equiv 6^n \pmod {100}$.
...
Third option.  $76*6^n = 50*6^n + 25*6^n + 6n^2 = 100(\frac {6^n}2+ \frac {6^n}4) + 6^n$. 
