Find the last three digits of $19^{100}$ Find the last three digits of $19^{100}$

$19^{100}=361^{50}=(1+360)^{50}=\binom{50}{0}+\binom{50}{1}360+\binom{50}{2}360^2+\cdots$
When we divide it by $1000$, the remainder comes out to be $001$, so the last three digits must be $001$, but in my book, the answer is given as $801$. I dont know where I am wrong. Please help.

Alternatively:
$$19^{100}=(20-1)^{100}=\underbrace{20^{100}-100\cdot 20^{99}+\cdots -\begin{pmatrix} 100\\ 3\end{pmatrix}\cdot 20^3}_{=1000A}+\begin{pmatrix} 100\\ 2\end{pmatrix}\cdot 20^2-\begin{pmatrix} 100 \\ 1\end{pmatrix}\cdot 20+\begin{pmatrix} 100 \\ 0\end{pmatrix}=$$
$$1000A+1978000+1.$$
Hence:
$$19^{100} \equiv 1 \pmod{1000}.$$
