7. Represent the number 531441 as the sum of several addends such that the first one is 1, the second is $k$ times larger, the third is $k$ times larger than the sum of the first two, the fourth is $k$ times larger than the sum of all previous ones, and so on. Find the multiplier $k$ and the number of addends, given that it is no less than four.
7. The condition of the problem can be written as:

$$
531441=3^{12}=1+k+(1+k) k=(1+k)^{2} k+\ldots+(1+k)^{n-3} k+(1+k)^{n-2} k
$$

Since the last term is $(1+k)^{n-2} k$, the sum of all previous terms is $(1+k)^{n-2}$. Therefore, $3^{12}=(1+k)^{n-2}+(1+k)^{n-2} k=(1+k)^{n-1}$. From this, $1+k=3^{\frac{12}{n-1}}$

The number of terms is no less than four. We select $n$, knowing that $\frac{12}{n-1}$ is an integer.

Answer: $k=80 ; 26 ; 8 ; 2 . n=4 ; 5 ; 7 ; 13$.