A finite sum involving the binomial coefficients and the harmonic numbers 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?
 A: 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.
A: 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 
$$\sum_{k=1}^n(-1)^{k-1}{n\choose k} \frac{H_k}{k}=H_n^{(2)}$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\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}
