Finding the last two digits of $6543^{210}$

One can use the binomial theorem (thrice). To do so, write $h$ for everything that is a multiple of $100$, possibly varying from line to line.

Since $6543=43+h$, one knows that $$ 6543^{210}=(43+h)^{210}=43^{210}+h. $$ Now, $$ 43^{210}=(3+40)^{210}=3^{210}+210\cdot3^{209}\cdot40+h=3^{210}+h. $$ Finally, $$ 3^{210}=9^{105}=(-1+10)^{105}=-1+105\cdot10+h=1049+h=49+h, $$ that is, $6543^{210}=49+$ some multiple of $100$.

This might be the fastest way: For the last two digits, you want to look modulo $100$. Notice that your number is relatively prime to $100$, and that $\phi(100)=40$. By Euler's theorem $$6543^{210}\equiv 6543^{10}\equiv 43^{10}\pmod{100}.$$ Here we can try using repeated squaring. $$43^2\equiv 49\pmod{100}.$$ $$43^4\equiv 49^2\equiv 1\pmod{100}.$$ Since $43^4\equiv 1$, we see that $$43^{10}\equiv 43^2\equiv 49\pmod{100}.$$

Binomial Theorem $\Rightarrow\rm mod\ 100\!:\ (-7\!+\!50)^{\large 2+4N}\!\equiv (-7)^{\large 2+4N}\! \equiv 7^{\large 2}\, $ by $\, 7^{\large 4}\! \equiv (-1\!+\!50)^{\large 2}\!\equiv 1$