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What's the remainder if $99{,}999^{99}$ is divided by $999{,}999$?

Is there any formula or trick method to achieve this?
Also kindly ignore the improper use of tag as i don't know which tag to choose

$!bmod 999999!:, 10cdot99999equiv -9 $ so $ 99999 equiv -9/10$

Thus $,99999^{large99}equiv -9^{large 99}/10^{large 99}equiv -9^{large 99}/10^{large 3}equiv -10^{large 3}cdot 9^{large 99} $ via $ 10^{large
6}equiv 1$

$n := 999999/27 = 7cdot 11cdot 13cdot 37.,$ mod each $p,,$ $9,$ has order $,3,5,3,9,$ so $,9^{large color{#c00}{45}}equiv 1pmod{!n}$

so $ 9^{large 99}!bmod 999999 = 27(3cdot 9^{large color{#c00}{97}}!bmod n) = 27(3cdot 9^{large 7}!bmod n) equiv 9^{large 9}!pmod{!999999}$

Hence we conclude $ 99999^{large 99}equiv -10^{large 3}cdot 9^{large 99}equiv {-}10^{large 3}cdot 9^{large 9}equiv 123579,pmod{!999999}$

We work in the ring $R=Bbb Z/N$, with operation modulo $N=999999=10^6-1$.
(Equalities below address computations in $R$.)
Then
$$
begin{aligned}
99999^{99}
&=(99999-999999)^{99} =(-900000)^{99}=-9^{99}cdot (10^5)^{99}\
&=-3^{198}cdot 10^{495}=-3^{198}cdot {underbrace{(10^6)}_{=1}}^{82}cdot 10^3=-3^{198}cdot 1000 .
end{aligned}
$$
Now the order of (the unit) $3$ in the ring

$Bbb Z/37037
=Bbb Z/(7cdot 11cdot 13cdot 37)
cong
(Bbb Z/7)times
(Bbb Z/11)times
(Bbb Z/13)times
(Bbb Z/37)
$

is $90$, but we need only the simpler information that $3^{180}$ is one modulo $37037$. This is so because $180$ is a multiple of $(7-1)$, $(11-1)$, $(13-1)$, and $(37-1)$. So instead of $3^{198}$ we write the smaller power $3^{18}=387420489$.

We finally search for a number which is $0$ modulo $27$, and also $-387420489000$ modulo $37037$, which is $12468$. After rearrangements modulo $3,9,27$ we get the result
$$123579 .$$

Note: A computer algebra system like sage delivers immediately

sage: Zmod(999999)(99999)^99

123579