show that ${n \choose 1}-\dots+(-1)^{n-1}{n \choose n} \left(1+\frac{1}{2}+\frac{1}{3}+\dots+\frac{1}{n}\right)=\frac{1}{n}$ Show that
${n \choose 1}$ - $ {n \choose 2}$ (1 + $ \frac{1}{2} $ )  ..............+ $(-1)^{n-1} {n \choose n}$ (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{n} $) = $ \frac{1}{n} $
MY ATTEMPT :  I tried by taking the general term
$T_r$ = $(-1)^{r-1}$ ${n \choose r}$  (1 + $ \frac{1}{2} $ + $ \frac{1}{3} $ +$ \frac{1}{4} $ ............+$ \frac{1}{r} $)
It looked some what like  ${n \choose r}$ $\int x^{r-1} $  =  ${n \choose r}$ $\frac {x^r}{r} $  so  I took an expansion say
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(1+ x+x^2+...........+x^{r-1})$  this simplifies to
$T_1(r)$ = $(-1)^{r-1}$ ${n \choose r}$ $(\frac{1-x^r}{1-x})$
I am struck after this I tried to rearrange this and done some simplification but I failed
Please help me

\begin{align}
\sum_{r=1}^n (-1)^{r-1} \binom{n}{r} \sum_{k=1}^r \frac{1}{k}
&= \sum_{k=1}^n \frac{1}{k} \sum_{r=k}^n (-1)^{r-1} \binom{n}{r} \\
&= \sum_{k=1}^n \frac{1}{k} (-1)^{k-1} \binom{n-1}{k-1} \\
&= \frac{1}{n}\sum_{k=1}^n (-1)^{k-1} \frac{n}{k} \binom{n-1}{k-1} \\
&= \frac{1}{n}\sum_{k=1}^n (-1)^{k-1} \binom{n}{k} \\
&= \frac{1}{n}
\end{align}
