$\sum_{k=1}^nH_k = (n+1)H_n-n$. 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?

Let $S_n$ denotes $\displaystyle\sum_{k=1}^n H_k$ and by applying Abel's summation:
$$\displaystyle\sum_{k=1}^n a_k b_k=A_nb_{n}+\sum_{k=1}^{n-1}A_k\left(b_k-b_{k+1}\right)\ $$ where $\ \displaystyle A_n=\sum_{i=1}^n a_i\ $ and letting $\ \displaystyle a_k=1 $ , $\ \displaystyle b_k=H_k\ $, we get,
\begin{align}
S_n&=\left(\sum_{i=1}^n1\right)H_{n}+\sum_{k=1}^{n-1}\left(\sum_{i=1}^k1\right)\left(H_k-H_{k+1}\right)\\
&=nH_{n}+\sum_{k=1}^{n-1}(k)\left(-\frac1{k+1}\right), \quad\color{blue}{\sum_{k=1}^{n-1}\frac{k}{k+1}=\sum_{k=0}^{n-1}\frac{k}{k+1}}\\
&=nH_{n}-\sum_{k=1}^n\frac{k-1}{k}\\
&=nH_{n}-\sum_{k=1}^n1+\sum_{k=1}^n\frac{1}{k}\\
&=nH_n-n+H_n\\
&=H_n(n+1)-n\\
\end{align}
