Wikipedia has a proof of the identity $$ H_{n} =\sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} \frac{1}{k}$$
http://en.wikipedia.org/wiki/Harmonic_number#Calculation
Curiously, there is also the identity
$$ \frac{1}{n} = \sum_{k=1}^{n} (-1)^{k-1} \binom{n}{k} H_{k} $$
Can the second identity be derived from the first one?
This is a consequence of binomial inversion. In general, if $f_{n}$ and $g_{n}$ are sequences such that $$f_{n} = \sum_{k=0}^{n}g_{k}{n \choose k}$$
then
$$g_{n} = \sum_{k=0}^{n}(-1)^{n+k}f_{k}{n\choose k}$$
Here I take sequences to be functions from $\mathbb{N} \rightarrow \mathbb{R}$ where I am using the convention that the natural numbers start at $0$.
The second of your identities can be proved from the first using binomial inversion by taking $$f_n = \left\{ \begin{array}{lr} 0 & : n = 0\\ H_{n} & : n > 0 \end{array} \right. $$ and $$g_n = \left\{ \begin{array}{lr} 0 & : n = 0\\ (-1)^{n-1}\frac{1}{n} & : n > 0 \end{array} \right. $$
Several proofs of binomial inversion are given in the answer I linked to.
We have
$$(x+y)^n=\sum_{k=0}^n{n\choose k}x^{n-k}y^k$$
set $x=1$ and $y=-t$ we get
$$(1-t)^n=\sum_{k=0}^n{n\choose k}(-1)^k t^k=-1+\sum_{k=1}^n(-1)^k{n\choose k} t^k$$
Differentiate both sides with respect to $t$
$$-n(1-t)^{n-1}=\sum_{k=1}^n(-1)^k{n\choose k} k\ t^{k-1}$$
Now multiply both sides by $\ln(1-t)$ then $\int_0^1 dt$ and use the fact that $\int_0^1 t^{k-1}\ln(1-t)\ dt=-\frac{H_k}{k}$
$$\sum_{k=1}^n(-1)^{k-1}{n\choose k} H_k=-n\int_0^1 (1-t)^{n-1}\ln(1-t)\ dt$$ $$\overset{1-t\to t}{=}-n\int_0^1 t^{n-1}\ln(t)\ dt=-n\left(-\frac{1}{n^2}\right)=\frac1n$$
Bonus:
If we follow the same approach without differentiating we will have
$$\sum_{k=1}^n(-1)^{k-1}{n\choose k} \frac{H_k}{k}=H_n^{(2)}$$
$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[5px,#ffd]{\sum_{k = 1}^{n}\pars{-1}^{k - 1}{n \choose k}H_{k}} = -\sum_{k = 1}^{n}H_{k}\,\pars{-1}^{k}{n \choose n - k} \\[5mm] = &\ -\sum_{k = 1}^{n}H_{k}\,\pars{-1}^{k}\bracks{z^{n - k}} \pars{1 + z}^{n} = -\bracks{z^{n}}\pars{1 + z}^{n}\sum_{k = 1}^{n}H_{k}\,\pars{-z}^{k} \\[5mm] = &\ \bracks{z^{n}}\pars{1 + z}^{n}\,{\ln\pars{1 + z} \over 1 + z} = \bracks{z^{n}}\pars{1 + z}^{n - 1}\bracks{\nu^{1}}\pars{1 + z}^{\nu} \\[5mm] = &\ \bracks{\nu^{1}}\bracks{z^{n}}\pars{1 + z}^{n + \nu - 1} = \bracks{\nu^{1}}{n + \nu - 1 \choose n} = \bracks{\nu^{1}} {\Gamma\pars{n + \nu} \over \Gamma\pars{n + 1}\Gamma\pars{\nu}} \\[5mm] = &\ \bracks{\nu^{1}}\nu\, {\Gamma\pars{n + \nu} \over \Gamma\pars{n + 1}\Gamma\pars{\nu + 1}} = \bracks{\nu^{0}} {\Gamma\pars{n + \nu} \over \Gamma\pars{n + 1}\Gamma\pars{\nu + 1}} \\[5mm] = &\ {\Gamma\pars{n} \over \Gamma\pars{n + 1}\Gamma\pars{1}} = \bbx{\large{1 \over n}} \\ & \end{align}
