7. (6 points) $n$ is a positive integer, and a prime number of the form $2^{n}-1$ is called a Mersenne number, for example: $2^{2}-1=3,2^{3}-1=7$ are Mersenne numbers. Recently, American scholars have refreshed the largest Mersenne number, $n=74207281$, and this Mersenne number is also the largest known prime number so far, its unit digit is . $\qquad$
【Analysis】According to the problem, this Mersenne number is $2^{n}-1=2^{74207281}-1$, to find the unit digit of the Mersenne number, we only need to find the unit digit of $2^{74207281}$, and the unit digit of $2^{74207281}$ can be determined according to the periodic rule.

【Solution】Solution: According to the analysis, this Mersenne number is $2^{n}-1=2^{74207281}-1$,
$\because 2^{1}=2 ; 2^{2}=4 ; 2^{3}=8 ; 2^{4}=16 ; 2^{5}=32$;
$2^{6}=64 ; 2^{7}=128 ; 2^{8}=256 ; 2^{9}=512 ; 2^{10}=1024 \cdots$
From this, we can see that the unit digit of $2^{n}$ is: $2, 4, 8, 6, 2, 4, 8, 6, 2 \cdots$
That is, when $n=1,5,9, \cdots$, the unit digit is $2 ; n=2,6,10, \cdots$, the unit digit is 4 ;
$n=3,7,11, \cdots$, the unit digit is $8 ; n=4,8,12, \cdots$, the unit digit is 6 ;
In summary, the unit digit of $2^{n}$ appears in a cycle, with a period of 4, and $74207281=4 \times 18551820+1$, so the unit digit of $2^{74207281}$ is the same as that of $2^{1}$, it can be concluded that the unit digit of $2^{74207281}$ is 2, and the unit digit of $2^{74207281}-1$ is: $2-1=1$.

Therefore, the answer is: 1.