Euler's Theorem states that if two positive integers $n$ and $a$ are co-primes then
$$a^{\varphi(n)} \equiv 1 \pmod n$$
Where $\varphi(n)$ is Euler Totient function.
Let $a=4444, n=9$ and $\varphi(9) = \varphi(3^2) = 3^2-3^1=6$ [ref] we get
$$(4444)^6 \bmod 9 = 1 \ \ \ \ \ \ \ \ \ \cdots(1)$$
Remember, $444 = 2^2 \times 11 \times 101$ and $9 = 3^2$ are co-primes.
We will also need multiplication property of modular arithmetic
$$(A \times B) \bmod C = (A \bmod C \times B \bmod C) \bmod C$$
Note, this property can be directly extended to exponents as exponentials are nothing but repeated multiplication.
Now we have enough resources to find the answer.
$$\begin{align}(4444)^{4444} \bmod 9 &=(4444)^{(6 \times 740 + 4)} \bmod 9 \\ &= \left[(4444^6)^{740} \times (4444)^4\right] \bmod 9 \\ &= \left[ (4444^6)^{740} \bmod 9 \ \times \ (4444)^4 \bmod 9\right] \bmod 9 \\ &= \left[(4444^6 \bmod 9)^{740} \bmod 9 \times \ (4444)^4 \bmod 9\right] \bmod 9 \\\end{align}$$
Use $(1)$ in the above result to get,
$$\begin{align}(4444)^{4444} \bmod 9 &= (4444)^4 \bmod 9 \\ &=\left[ 4444 \bmod 9\right]^4 \bmod 9 \\ &= 7^4 \bmod 9 \\ &= \left[ 49 \bmod 9 \times 49 \bmod 9\right] \bmod 9 \\ &= 16 \bmod 9 \\ &= 7\end{align}\\$$
There is little advantage we get using Euler's Theorem. We can hit this problem without it, using just multiplication property of modular arithmetic.
Here something hard to identify initially, but true is $7^3 \equiv 1 \pmod 9$ $\therefore (7^3)^{1481} \equiv 1^{1481} \pmod 9 $
$$\begin{align}(4444)^{4444} \bmod 9 &= \left( 4444 \bmod 9 \right)^{4444} \bmod 9 \\ &= 7^{4444} \bmod 9 \\ &= \left[ 7^{4443} \times 7 \right] \bmod 9 \\ &= \left[ (7^3)^{1481} \times 7 \right] \bmod 9 \\ &= 7 \bmod 9 \\ &= 7\end{align}$$
HTH