$\sum_{k=1}^nH_k = (n+1)H_n-n$. Why? This is motivated by my answer to this question.  
The Wikipedia entry on harmonic numbers gives the following identity: 
$$
\sum_{k=1}^nH_k=(n+1)H_n-n
$$
Why is this?  
Note that I don't just want a proof of this fact (It's very easily done by induction, for example).  Instead, I want to know if anyone's got a really nice interpretation of this result: a very simple way to show not just that this relation is true, but why it is true.  
Has anyone got a way of showing that this identity is not just true, but obvious?

I suck at making pictures, but I try nevertheless. Write $n+1$ rows of the sum $H_n$:
$$\begin{matrix}
1 & \frac12 & \frac13 & \dotsb & \frac1n\\
\overline{1\Big\vert} & \frac12 & \frac13 & \dotsb & \frac1n\\
1 & \overline{\frac12\Big\vert} & \frac13 & \dotsb & \frac1n\\
1 & \frac12 & \overline{\frac13\Big\vert}\\
\vdots & & &\ddots & \vdots\\
1 & \frac12 &\frac13 & \dotsb & \overline{\frac1n\Big\vert}
\end{matrix}$$
The total sum is obviously $(n+1)H_n$. The part below the diagonal is obviously $\sum\limits_{k=1}^n H_k$. The part above (and including) the diagonal is obviously $\sum_{k=1}^n k\cdot\frac1k = n$.
It boils down of course to the same argument as Raymond Manzoni gave, but maybe the picture makes it even more obvious.
