Last $4$ digits of $7^{128}$? Trying to find the easiest and most general method in finding last $n$ digits of a number.
I know the trick lies in finding the remainder when the number is divided by $10^n$ but still not able to perform the rest of the steps to reach the answer
 A: Repeated squaring:
$$7^4 = 2401 \mod 10000$$
$$7^8 = 2401^2\equiv 4801 \mod 10000$$
$$7^{16}\equiv 4801^2\equiv 9601 \mod 10000$$
$$7^{32}\equiv 9201 \mod 10000$$
$$7^{64}\equiv 8401 \mod 10000$$
$$7^{128}\equiv 6801\mod 10000$$
Note:  $2401^2=(2400+1)^2=2400^2+4800+1\equiv4801$, $4801^2\equiv4800^2+9600+1\equiv9601$, 
$9601^2\equiv9600^2+2\times9600+1\equiv9200,$ etc.
A: $7^{128}\equiv(1-50)^{64}\equiv1-64\cdot50+\binom{64}250^2-\binom{64}350^3+\cdots+50^{64}$
$\equiv1-64\cdot50\pmod{10^4}$
as $10^4$ divides $50^n,n\ge4,$
and $10^4\mid\binom{64}250^2\iff\dfrac{10^4}{50^2}\mid\binom{64}2$
and similarly $10^4$ divides $\binom{64}350^3$
A: Below I show that the method in JWT's answer is also essentially Binomial Theorem based.
$\begin{align} 
7^{\large 2}\ &= -1 + 50 \\[.3em]
\Rightarrow\ \ \ \ \ \ \ 7^{\large 4}\ &=\ \ \  1 + 24\cdot100\ \, (=\, 1  -2(50)+ 2500\ \ \rm by\ squaring\ above)\\[.3em]
\Rightarrow\  (7^{\large 4})^{\large 32}  &\equiv \ \ \ 1 + \color{#0a0}{32\cdot 24}\cdot 100\!\!\pmod{\!100^{\large 2}}\ \ \rm by\ Binomial\ Theorem\\[.3em]
                          &\equiv\ \ \  1\  \, +\, \  \color{#c00}{68}\,\ \cdot\,\ 100\ \ \ {\rm by}\ \  \color{#0a0}4\underbrace{(\color{#0a0}{8\cdot 24}\bmod 25)}_{\large 8\,(-1)\ \,\equiv\,\ \color{#c00}{17}\  \ \ \ \ \ \ } = \color{#c00}{4(17)}
\end{align}$ 
