An interesting sum identity: $\sum \frac 1 {k_1\cdots k_m} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k^m} \binom n k$ Fix a positive integer $m$.  I am trying to prove
\[\sum \frac 1 {k_1\cdots k_m} = \sum_{k=1}^n \frac{(-1)^{k-1}}{k^m} \binom n k\]
where the sum on the left is taken over all integer $m$-tuples $(k_1,\dots,k_m)$ such that $1\le k_1\le \cdots\le k_m\le n$.  This comes immediately after having proved that
\[ \sum_{k=1}^n b_k/k = \sum_{k=1}^n \frac{a_k}k \binom n k \]
where $b_n = \sum_{k=1}^n a_k\binom n k$.  There's a strong similarity, so I suspect that I should use this result to prove the above.  In particular it suggests that we take $a_k/k = (-1)^{k-1}/k^m$ which is to say $a_k = (-1)^{k-1}/k^{m-1}$.  If this is the right path then we want to show that
\[ \sum\frac 1 {k_1\cdots k_m} = \sum_{k=1}^n \frac{b_k}k = \sum_{k=1}^n \frac 1 k \sum_{j=1}^k a_j\binom{k}j = \sum_{k=1}^n \frac 1 k \sum_{j=1}^k \frac{(-1)^{j-1}}{j^{m-1}}\binom{k}j\]
Now trying to rearrange and reindex the sums basically just seems to get you back into the proof that $\sum b_k/k = \sum a_k\binom n k /k$ so that seems like a dead end.
I'm guessing that somehow, instead, we should find a more direct reason why these are equal, perhaps by splitting fractions.  For instance, $\frac 1 {2\cdot 3} = \frac{1}{2}-\frac{1}{3}$ but there is no such simple splitting for $\frac 1 {2\cdot 4}$.
The right-hand side of the above seems to suggest that perhaps we fix a $1\le k \le n$ as the least number in $(k_1,\dots,k_m)$, and then use something that looks vaguely like inclusion-exclusion.
From here the question looks like "Is there a reason why, if we collect all the terms of $\sum \frac 1 {k_1\cdots k_m}$ with a fixed $k_1$, that the sum of these terms is $\frac 1 k \sum_{j=1}^k \frac{(-1)^{j-1}}{j^m}\binom k j$?"
Just to test the idea out, if we set $n=4,m=2$ and $k_1=2$ then we would be looking at the terms
\[ \frac{1}{2\cdot 2}+\frac{1}{2\cdot 3}+\frac{1}{2\cdot 4} \]
and on the other hand we would be trying to see whether this is the same as
\[ \frac 1 2 \sum_{j=1}^2 \frac{(-1)^{j-1}}{j^2}\binom 2 j = \frac 1 2 \cdot \left( 2 - \frac 1 4  \right)\]
A quick computation shows these are not equal, so then I'm not sure if I made a mistake along the way, or if the entire plan of attack is flawed.
If the entire plan is flawed, I don't have a plan B.
 A: Let $S_{m,n}$ be your first sum, so
\begin{equation}
\begin{split}
S_{m+1,n}&=\sum_{1\leq k_1\leq\cdots\leq k_{m+1}\leq n}\frac{1}{k_1k_2\cdots k_mk_{m+1}}\\
&=\sum_{\ell=1}^n\left[\sum_{1\leq k_1\leq\cdots \leq k_m\leq\ell}\frac{1}{k_1k_2\cdots k_m\ell}\right]\\
&=\sum_{\ell=1}^n\frac{1}{\ell}S_{m,\ell}\\
\end{split}
\end{equation}
Now, we proceed by induction on $m$. Suppose that
\begin{equation}
S_{m,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}{n\choose k}
\end{equation}
for all $n\geq 0$, then
\begin{equation}
\begin{split}
S_{m+1,n}&=\sum_{\ell=1}^n\frac{1}{\ell}S_{m,\ell}\\
&=\sum_{\ell=1}^n\frac{1}{\ell}\sum_{k=1}^\ell\frac{(-1)^{k-1}}{k^m}{\ell\choose k}\\
&=\sum_{\ell=1}^n\sum_{k=1}^\ell\frac{(-1)^{k-1}}{\ell k^m}{\ell\choose k}\\
\end{split}
\end{equation}
Changing the order of summation gives us
\begin{equation}
\begin{split}
S_{m+1,n}&=\sum_{k=1}^n\sum_{\ell=k}^n\frac{(-1)^{k-1}}{\ell k^m}{\ell\choose k}\\
&=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\sum_{\ell=k}^n\frac{1}{\ell}{\ell\choose k}\\
&=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}\sum_{\ell=k}^n\frac{1}{k}{{\ell-1}\choose {k-1}}\\
&=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^{m+1}}\sum_{\ell=k}^n{{\ell-1}\choose {k-1}}\\
\end{split}
\end{equation}
Finally, using a modified version of the hockey-stick identity $\sum_{\ell=k}^n{{\ell-1}\choose {k-1}}={n\choose k}$ (when $n\geq k$), we have that
\begin{equation}
S_{m+1,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^{m+1}}{n\choose k}\\
\end{equation}
which finishes our inductive step. Now, note that there exists one tuple of length $0$, and the empty product is defined as $1$, so by definition
\begin{equation}
S_{0,n}=1=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^0}{n\choose k}
\end{equation}
where the second equality is true by the binomial theorem, which proves the base case. Therefore, by induction
\begin{equation}
S_{m,n}=\sum_{k=1}^n\frac{(-1)^{k-1}}{k^m}{n\choose k}
\end{equation}
for all $m,n\geq 0$.
