How to find $\sum_{A \subset S} (\min A)$ and $\sum_{A \subset S} (\max A)$ if $S=\{1,2,...,n\}$?

Note. I assume that $m(\emptyset)=0$.

Question. Consider the set $S=\{1,2,3,\dots,j\}$. Let $m(A)$ denote a maximum element of a subset $A$ of $S$. Prove that $$\sum_{A\subseteq S}m(A) = (j-1)2^j+1$$ where summation ranges over all subsets $A$ of $S$.

Solution. Let us consider the element $4$. $4$ will be maximum element if we pick elements from the set $\{1,2,3,4\}$. So basically we want to count number of subsets of the set $\{1,2,3,4\}$ which contain the element $4$. This is the same as counting the number of subsets of $\{1,2,3\}$ and for each subset of $\{1,2,3\}$ we append the element $4$ to it. There are $2^3$ subsets of $\{1,2,3\}$, namely \begin{align*} \binom30 &= \{\emptyset\} \to \{4\}=\{4\}\\ \binom31 &= \{\{1\},\{2\},\{3\}\} \to \{4\}=\{\{1,4\},\{2,4\},\{3,4\}\}\\ \binom32 &= \{\{1,2\},\{1,3\},\{2,3\}\} \to \{4\}=\{\{1,2,4\},\{1,3,4\},\{2,3,4\}\}\\ \binom33 &= \{1,2,3\} \to 4=\{1,2,3,4\} \end{align*} So this gives us a one-one correspondence between the subsets of $\{1,2,3,4\}$ containing $4$ and the subset of $\{1,2,3\}$. So in general the subset of $\{1,2,3,\dots,j\}$ containing $j$ is in one-one correspondence between the subsets of $\{1,2,3,\dots,(j-1)\}$. Therefore the sum $$\sum_{A\subseteq S}m(A) = \sum_{i=1}^j i \cdot 2^{i-1}= (j-1)2^j+1.$$ Note $1\cdot1+2\cdot2+3\cdot2^3+4\cdot2^3+\dots+n\cdot2^{n-1}= \frac{d}{dx} (x+x^2+x^3+\dots+x^n)_{x=2}$