7. A number table composed of several rows of numbers, starting from the second row, each number in the row is equal to the sum of the two numbers above it, the last row contains only one number, the first row is the first 100 positive integers arranged in ascending order. Then the number in the last row is (exponential notation can be used).


7. $101 \times 2^{98}$.

It is easy to know:
(1) The table has a total of 100 rows;
(2) Each row forms an arithmetic sequence, and the common differences are
$$
d_{1}=1, d_{2}=2, d_{3}=2^{2}, \cdots, d_{99}=2^{98} ;
$$
(3) Let the first number of the $n(n \geqslant 2)$-th row be $a_{n}$. Then $a_{100}$ is the desired value.
$$
\begin{array}{l}
a_{n}=a_{n-1}+\left(a_{n-1}+2^{n-2}\right)=2 a_{n-1}+2^{n-2} \\
=2\left(2 a_{n-2}+2^{n-3}\right)+2^{n-2} \\
=2^{2}\left(2 a_{n-3}+2^{n-4}\right)+2 \times 2^{n-2}+2^{n-2} \\
=2^{3} a_{n-3}+3 \times 2^{n-2}=\cdots \\
=2^{n-1} a_{1}+(n-1) \times 2^{n-2} \\
=(n+1) 2^{n-2} .
\end{array}
$$

Therefore, $a_{100}=101 \times 2^{98}$.