7. Let set $S$ contain $n$ elements, $A_{1}, A_{2}, \cdots, A_{k}$ are different subsets of $S$, they all have non-empty intersections with each other, and no other subset of $S$ can intersect with all of $A_{1}, A_{2}, \cdots, A_{k}$. Express this in terms of $n$:


7. $k=2^{n-1}$.

If $k>2^{n-1}$, since $S$ has $2^{n}$ subsets, they can be paired into $2^{n-1}$ pairs, with each pair consisting of two complementary subsets. By $k>2^{n-1}$, there are two complementary subsets among $A_{1}, A_{2}, \cdots, A_{k}$, thus their intersection is an empty set, which is a contradiction; if $k<2^{n-1}$, from $S$ excluding $A_{1}, A_{2}, \cdots, A_{k}$, select a pair of complementary subsets $X$ and $Y$. According to the problem, there is an $A_{i}$ among $A_{1}, A_{2}, \cdots, A_{k}$ such that $A_{i} \cap X=\varnothing$, thus $A_{i} \subset Y$, and there is an $A_{j}$ such that $A_{j} \cap Y=\varnothing$. Then $A_{j} \subset X$. Therefore, $A_{j} \cap A_{j} \subset X \cap Y=\varnothing$, i.e., $A_{i} \cap A_{j}=\varnothing$, which is a contradiction. Thus, $k=2^{n-1}$.