By The Fermat-Euler Theorem we know ${ 21 }^{ 400 }\equiv 1\quad mod\quad 1000$ , so we only have to find ${ 21 }^{ 168 }$ .

Now, see ${ 21 }^{ 5 }\equiv 101\quad mod\quad 1000$ , applying it in the last equation $101^{ 32 }*2{ 1 }^{ 8 }\quad mod\quad 1000$ . See, $101^{ 2 }\equiv 201,\quad 101^{ 3 }\equiv 301,\quad ...,\quad 10{ 1 }^{ 10 }\equiv 1\quad mod\quad 1000$ . So we only have to find $101^{ 2 }*2{ 1 }^{ 8 }\equiv 201*361\equiv \boxed { 561 } \quad mod\quad 1000$ .

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prod = 1
for i in range(1,569):
    prod *= 21
    prod %= 1000
print prod 

JuanTamad