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

First, let's summarize what you've already shown:  The given sum is equivalent to
$$\sum_{r=1}^n \int_0^1\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx$$
The first thing we'll do to this is make the sum start at $r=0$ instead of $r=1$ (which doesn't change anything since when $r=0$ the integrand is $0$), move the [finite] sum inside the integral, and then rearrange algebraically:
$$\begin{align*}\sum_{r=1}^n \int_0^1&\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx = \sum_{r=0}^n \int_0^1\left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx \\ &= \int_0^1 \sum_{r=0}^n \left((-1)^{r-1}\binom{n}{r}\frac{1-x^r}{1-x}\right)\,dx \\ &= \int_0^1 (1-x)^{-1}\left(\sum_{r=0}^n\binom{n}{r}(-x)^r-\sum_{r=0}^n\binom{n}{r}(-1)^r\right)\,dx\end{align*}$$
Now both sums have the form $\sum_{r=0}^n \binom{n}{r} a^r = (1+a)^n$ by the binomial theorem, so by simplifying:
$$\int_0^1(1-x)^{-1}\left((1-x)^n - (1-1)^n\right)\,dx = \int_0^1 (1-x)^{n-1}\,dx = \frac{1}{n}$$
