$5^{1234}$ is not too large

Since $\left\{\big(a*b\big) \bmod c =\Big(\Big(\big(a \bmod c\big)*\big(b \bmod c\big)\Big) \bmod c\Big)\right\}$

And ${5^6} \bmod 9 = 1$ so, for any positive integer $k$ , ${5^{6k}} \bmod 9 = 1$ ,

because

${\Big(5^6*5^6*...*5^6\Big)} \bmod 9=\Big(\big(5^6 \bmod 9\big) * \big(5^6 \bmod 9 \big)* ... * \big(5^6 \bmod 9\big)\Big) \bmod 9 = \big(1*1*...*1\big) \bmod 9 = 1$ .

$5^{1230}=5^{6*205}$ , so , $5^{1230} \bmod 9 =1$

${5^{1234}} \bmod 9 = \Big(\big(5^{1230} \bmod 9\big) * \big(5^2 \bmod 9 \big)* \big(5^2 \bmod 9\big)\Big) \bmod 9 = \big(1*7*7\big) \bmod 9 = 49 \bmod 9 = 4$

Relevant wiki: Euler's Theorem

Since $5$ and $9$ are relatively prime, by Euler's theorem $\begin{aligned} 5^{1234} \bmod 9 &\equiv 5^{1234\, \bmod\, \phi(9)} &\mod 9 \\ &\equiv 5^{1234\, \bmod \, 6} &\mod 9 &&& \color{#3D99F6} \phi(9)=6 \\ &\equiv 5^4 & \mod 9 &&& \color{#3D99F6} 1236 \equiv 0\bmod 6 \text{, so } 1234 \equiv 4\bmod 6 \\ &\equiv 625 & \mod 9 \\ &\equiv \boxed{4} & \mod 9 &&& \color{#3D99F6} 6+2+5 = 13 \text{ and } 1 + 3 = 4 \end{aligned}$

By Euler's Theorem, $a^{\phi(n)} \equiv$ 1 mod n $\forall a, n \in N$ such that gcd $(a, n) = 1$ .

Since $9 = 3^2 \Rightarrow \phi(9) = 9(1 - \frac{1}{3}) = 9(\frac{2}{3}$ ) = 6 by definition of Euler's Totient Function.

$\therefore$ $5^{6k} \equiv$ 1 mod 9 $\forall$ integers $k \geq 1$ .

Since $1234 \equiv$ 4 mod 9 $\Rightarrow 5^{1234} \equiv 5^4$ mod 9 $\equiv 625$ mod 9.

$\therefore 5^{1234} \equiv 4$ mod 9 $\Rightarrow$ the remainder is $\boxed{4}$ .

Relevant wiki: Euler's Theorem

$\begin{aligned} 5^{1234} & \equiv 5^{\color{#3D99F6}1234 \bmod \phi (9)} \text{ (mod 9)} & \small \color{#3D99F6} \text{Since }\gcd(5,9) = 1 \text{, Euler's theorem applies.} \\ & \equiv 5^{\color{#3D99F6}1234 \bmod 6} \text{ (mod 9)} & \small \color{#3D99F6} \text{Euler's totient function }\phi (9) = 9 \times \frac 23 = 6 \\ & \equiv 5^4 \text{ (mod 9)} \\ & \equiv 25 \times 25 \text{ (mod 9)} \\ & \equiv 7 \times 7 \text{ (mod 9)} \\ & \equiv 49 \equiv \boxed 4 \text{ (mod 9)} \end{aligned}$