To the power of 2000

$(1991)^{2000}$

$=(2000-9)^{2000}$

$=2000^{2000} + \ldots +{2000 \choose 1}\cdot 2000\cdot (-9)^{1999} + (-9)^{2000}$

Since all the terms in this expansion except the last term are greater than $10^6$ , they will not contribute to the remainder. Hence:

$1991^{2000} \equiv 9^{2000} \pmod{10^6}$

Writing the binomial expansion again, we get:

$9^{2000} = (10-1)^{2000}$

$= 10^{2000} + \ldots + {2000 \choose 2}\cdot 10^{2} \cdot (-1)^{1998} + {2000 \choose 1}\cdot 10 \cdot (-1)^{1999} + (-1)^{2000}$

All the terms in the above expansion except the last three are greater than $10^6$ . Hence:

$1991^{2000} \equiv 199,900,000 - 20,000 + 1 \equiv \boxed{880,001} \pmod{10^6}$