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The problem can be expressed as ${1234}^{4321}\pmod{4321}$

Since $~~4321=29\times{149}$ and $29$ and $149$ are prime numbers, we can use Fermat's Little Theorem and Chinese Remainder Theorem to find out the final result. But, first, we should change the expression above becomes : ${1234}^{4321}\equiv\begin{cases}{1234}^{4321}\pmod{29}\\{1234}^{4321}\pmod{149}\end{cases}$

Now, using Fermat's Little Theorem we have ${1234}^{4321}\equiv\begin{cases}{1234}^{4321}\equiv\left({1234}^{28}\right)^{154}\times{{1234}^{9}}\equiv1\times{{1234}^{9}}\pmod{29}\\{1234}^{4321}\equiv\left({1234}^{148}\right)^{29}\times{{1234}^{29}}\equiv1\times{{1234}^{29}}\pmod{149}\end{cases}$

then, ${1234}^{4321}\equiv\begin{cases}{1234}^{9}\equiv{16}^{9}\equiv24\pmod{29}\\{1234}^{29}\equiv{42}^{29}\equiv130\pmod{149}\end{cases}$

Combine these result by using Chines Reminder Theorem to find the final result

$\begin{aligned}{1234}^{4321}&\equiv24\times{149}\times{22}+130\times{29}\times{360}\\&\equiv214392\equiv\boxed{2663}\pmod{4321}\end{aligned}$