First, $2003^n \equiv 3^n \mod 1000$.
$3$ is invertible modulo $1000$. The group of invertibles of $\mathbb{Z}/1000\mathbb{Z}$, $(\mathbb{Z}/1000\mathbb{Z})^\times$ has cardinality $\varphi(1000) = 1000 * 1/2 * 4/5 = 400$.
This implies that $3^{400} \equiv 1 \mod 1000$, and so $3^n \equiv 3^{n \mod 400} \mod 1000$.
So in order to comptute $2003^{2002^{2001}} \mod 1000$, we need to know $2002^{2001} \mod 400$.
$2002^n \equiv 2^n \mod 400$. This time, $2$ is not invertible modulo $400 = 2^4 * 25$.
For $n \geq 4$, $2^n$ is always a multiple of $2^4$, so $2^n \mod 400 = (2^4*2^{n-4}) \mod (2^4*25) = 2^4*(2^{n-4} \mod 25)$.
Now, $2$ is invertible modulo 25, and the group $(\mathbb{Z}/25\mathbb{Z})^\times$ has cardinality $\varphi(25) = 25*4/5 = 20$. This implies that $2^{20} \equiv 1 \mod 25$, and so $2^n \equiv 2^{n \mod 20} \mod 25$.
Putting all of this together, we get :
$2002^{2001} \mod 400 = 2^{2001} \mod 400 = 2^4 * (2^{1997} \mod 25) = 2^4 * (2^{1997 \mod 20} \mod 25) $
$=2^4 * (2^{17} \mod 25) = 2^4 * (131072 \mod 25) = 2^4 * 22 = 352$.
And finally $2003^{2002^{2001}} \mod 1000 = 3^{352} \mod 1000 = 241$.
As $\displaystyle2017\equiv17\pmod{1000},2017^m\equiv17^m$ for any integer $m$
Use Carmichael function, $\lambda(1000)=100$
$$\implies17^{(2016^{2015})}\equiv17^{(2016^{2015})\pmod{100}}\pmod{1000}$$
Now $2016\equiv16\implies2016^{2015}\equiv16^{2015}\pmod{100}$
As $(16,100)=4$ let use find $16^{2015-1}\pmod{100/4}$
$16^{2014}=(2^4)^{2014}=2^{8056}$
As $\displaystyle\lambda(25)=\phi(25)=20$ and $8056\equiv16\pmod{20},2^{8056}\equiv2^{16}\pmod{25}$
$2^8=256\equiv6\pmod{25}\implies2^{16}\equiv6^2\equiv11$
$\implies16^{2014}\equiv11\pmod{25}$
$\implies16^{2014+1}\equiv11\cdot16\pmod{25\cdot16}\equiv176\pmod{400}\equiv76\pmod{100}$
$$\implies17^{(2016^{2015})}\equiv17^{76}\pmod{1000}$$
Now $$17^{76}=(290-1)^{38}=(1-290)^{38}\equiv1-\binom{38}1290+\binom{38}2290^2\pmod{1000}$$
Now $\displaystyle38\cdot29=(40-2)(30-1)\equiv2\pmod{100}\implies\binom{38}1290\equiv20\pmod{1000}$
and $\displaystyle\binom{38}229^2=\dfrac{38\cdot37}2(30-1)^2\equiv3\pmod{10}$
$\displaystyle\implies\binom{38}2290^2\equiv3\cdot100\pmod{10\cdot100}$
Best Answer
It's $-16-15-14-...-1=-16\times17/2=-8\times17=-136\equiv1881\bmod2017$.