Too Many 4s

From Euler's theorem, $\begin{array}{l} 4444^{4444}\mod9 \\=(4444\mod9)^{4444\mod\phi(9)}\mod9 \\=(16\mod9)^{4444\mod6}\mod9 \\=7^4\mod9 \\=49^2\mod9 \\=4^2\mod9 \\=16\mod9 \\=7 \end{array}$

First note that $4444^{4444} = (9 * 493 + 7)^{4444} \equiv 7^{4444} \pmod{9}$ .

Next, note that $7^{3} = 343 = 9 * 38 + 1$ , and so $7^{3} \equiv 1 \pmod{9}$ .

Finally, note that $4444 = 3 * 1481+ 1$ , and so

$4444^{4444} \equiv 7^{3 * 1481 + 1} \equiv 1*7 \pmod{9}$ ,

i.e., the desired remainder is $\boxed{7}$ .