Vrijeme: 23:17

Nepotrebno bezobrazni | Unnecessarily rude #3

Tetracija je ponovljeno potenciranje i označava se ovako: ^{n}{a} = a^{a^{\dots^{a}}} gdje se a na desnoj strani pojavi točno n puta.
Nadalje, takve potencije se računaju odozgo prema dolje. Na primjer, ^{3}{3} = 3^{3^3} = 3^{\left( 3^3 \right) } = 3^{27} = 7625597484987
i ^{4}{2} = 2^{2^{2^2}} = 2^{\left( 2^{ \left( 2^2 \right) } \right) } = 2^{\left(2^4 \right) } = 2^{16} = 65536. Odredi:
 
 

^{1}{1} \;\cdot\; ^{2}{2} \;\cdot\; ^{3}{3} \;\cdot\; ^{4}{4} \;\cdot\; ... \;\cdot\; ^{111}{111} \pmod{11111111111111}

Tetration is repeated exponentiation and it is denoted like so: ^{n}{a} = a^{a^{\dots^{a}}} where a appears exactly n times on the right.
Furthermore, they are calculated from top to bottom. For example, ^{3}{3} = 3^{3^3} = 3^{\left( 3^3 \right) } = 3^{27} = 7625597484987
and ^{4}{2} = 2^{2^{2^2}} = 2^{\left( 2^{ \left( 2^2 \right) } \right) } = 2^{\left(2^4 \right) } = 2^{16} = 65536. Determine:
 
 

^{1}{1} \;\cdot\; ^{2}{2} \;\cdot\; ^{3}{3} \;\cdot\; ^{4}{4} \;\cdot\; ... \;\cdot\; ^{111}{111} \pmod{11111111111111}