8. Let $X_{n}=\{1,2, \cdots, n\}$, for any subset $A$ of $X_{n}$, denote $f(A)$ as the smallest element in $A$, then $\sum_{A \subseteq X_{n}} f(A)=$ $\qquad$ .
8. $2^{n+1}-2-n$ Detailed Explanation: Classify by the value of $f(A)$: Let $f(A)=i(i=1,2, \cdots, n)$, then $\sum_{A \leq X_{n}} f(A)=$ $\sum_{i=1}^{n} \sum_{f(A)=i, A \subseteq x_{n}} i$. When $f(A)=i$, $A$ does not contain any number smaller than $i$, and numbers larger than $i$ can either belong to $A$ or not, each having two possibilities. Therefore, there are $2^{n-i}$ such subsets $A$. Thus, $\sum_{A \subseteq X_{n}} f(A)=\sum_{i=1}^{n} \sum_{f(A)=i, A \subseteq x_{n}} i=\sum_{i=1}^{n} i \cdot 2^{n-i}$, and using the method of misalignment subtraction, we get $\sum_{A \leq X_{n}} f(A)=2^{n+1}-2-n$.