Digis:Find it out

using binomial....clearly....
9¹⁰⁰≡1(mod 1000)
now i think u can proceed further...

no, Ketan,, i got stuck.. Please proceed further...

\hspace{-16}\mathbb{F}$or the Calculation of Last $\bf{3}$ Digit in $\bf{9^{9^{9^{9}}}}$, First We Will Calculate $\bf{9^9}$ of\\\\ extreme Right Upper Part.\\\\ So $\bf{9^9=(10-1)^{9}=\binom{9}{0}.(10)^{10}-\binom{9}{1}.(10)^{9}+.....+\binom{9}{8}.(10)-\binom{9}{9}}$\\\\\\ So $\bf{9^9=\mathbb{M}(100)+90-1=89}$\\\\\\ Means Last $\bf{2}$ Digit of $\bf{9^9}$ is $\bf{=89}$, Where $\bf{\mathbb{M(\bf{100})}=M}$ultiple of $\bf{100}$\\\\\\ Now We Will Calculate $\bf{9^{89}}$\\\\\\ So $\bf{9^{89}=(10-1)^{89}=\binom{89}{0}.(10)^{10}-\binom{89}{1}.(10)^{9}+.......-\binom{89}{87}.(10)^2+\binom{89}{88}.(10)-\binom{89}{89}}$\\\\\\ So $\bf{9^{89}=\mathbb{M}(1000)-391600+890-1=\mathbb{M}(1000)+289}$\\\\\\ So Last $\bf{3}$ Digit of $\bf{9^{89}}$ is $\bf{289}$\\\\\\ Now Calculate $\bf{9^{289}}$\\\\\\ So $\bf{9^{289}=(10-1)^{289}=\binom{289}{0}.(10)^{10}-\binom{289}{1}.(10)^{9}+.......-\binom{289}{287}.(10)^2+\binom{289}{288}.(10)-\binom{289}{289}}$\\\\\\

\hspace{-16}$So $\bf{9^{289}=\mathbb{M}(1000)-4161600+2890-1=\mathbb{M(\bf{1000})+289}}$\\\\\\ So Last $\bf{3}$ Digit of $\bf{9^{9^{9^{9}}}}$is $\bf{\underline{\underline{289}}}$