a) $$\sum_{k=1}^{m}k {n+k\brace k} ={ n+m+1 \brace m}$$
It seems that we are supposed to check the number of divisions of $[n+k]$ into $k$ sets and then pick one of those sets. However, I'm at a loss since I have no clue what the $k$'s are supposed to count and my intuition is silent.
b) $$\sum_{k=1}^{m}(n+k) {n+k \brack k}={n+m+1 \brack m}$$
A very similar question with exactly the same issue for me. Here we divide take a permutation of $[n+k]$ with $k$ cycles and then pick any element from $[n+k]$, but again, what are the $k$'s counting?
Any hints would be appreciated.
For the first one, let $\pi$ be a partition of $[n+m+1]$ into $m$ blocks. Let $A(\pi)$ be the number such that $$(n+m-A(\pi)+1),(n+m-A(\pi)+2),\cdots ,(n+m+1)$$ are all singletons in $\pi$ (So there are at least $A(\pi)$ singletons) but $n+m-A(\pi)$ is not in a singleton (so we have to place it in $m-A(\pi)$ blocks). Then by doing $i=A(\pi)$ and adding over all possible $i$'s, we get $${n+m+1\brace m}=\sum _{i=0}^m{n+m-i \brace m-i}(m-i),$$ doing $k=m-i$ (so $k$ is the number of blocks on the partition that are not the like this last singletons) we get $${n+m+1\brace m}=\sum _{k=1}^{m}{n+k\brace k}k.$$
The second one is be pretty similar, the only change is that you have to place $n+m-A(\pi)$ in an ordered fashion.