Prove $\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \frac{1}{n+1} $ I am going through a proof of the combinatorial result $$\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \frac{1}{n+1}, $$ and have found something I don't understand.
The proof is as follows:
$$\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} = \sum_{k=0}^{n}{n \choose k}(-1)^k \int_0^1 t^k\ dt = \int_0^1 (1-t)^n\ dt = \frac{1}{n+1}.$$
I understand this, except from the second to last equality. Why does this equality hold?
Any help would be appreciated!
 A: An alternative proof:
$$\begin{align}
\sum_{k=0}^{n}{n \choose k}(-1)^k \frac{1}{k+1} 
&=\frac 1{n+1}\sum_{k=0}^{n}(-1)^k\frac {n+1}{k+1}{n \choose k} \\
&=\frac 1{n+1}\sum_{k=0}^n (-1)^k\binom {n+1}{k+1}\\
&=\frac 1{n+1}\sum_{k=1}^{n+1} (-1)^{k-1}\binom {n+1}{k}\\
&=\frac 1{n+1}\bigg[\underbrace{\sum_{k=0}^{n+1} (-1)^{k-1}\binom {n+1}{k}}_{=0}-(-1)\underbrace{\binom{n+1}0}_{=1}\bigg]\\
&=\frac 1{n+1}\qquad\blacksquare\end{align}$$
A: It’s also possible to give a combinatorial proof.
Let $A=\{0\}\cup[n]=\{0,\ldots,n\}$; we’ll count in two ways the permutations of $A$ that begin with $0$.
On the one hand there are of course $n!$ of them, since the remaining $n$ elements can be arranged in any order.
Alternatively, for $k\in[n]$ let $P_k$ be the set of permutations of $A$ in which $k$ precedes $0$; clearly exactly half of the permutations of $A$ are in $P_k$, so $|P_k|=\frac12(n+1)!$. More generally, if $\varnothing\ne I\subseteq[n]$, then $\bigcap_{k\in I}P_k$ is the set of permutations in which $0$ is preceded by every element of $I$, and
$$\left|\bigcap_{k\in I}P_k\right|=\frac{(n+1)!}{k+1}\;.$$
A straightforward inclusion-exclusion argument now yields the equation
$$\left|\bigcup_{k\in[n]}P_k\right|=\sum_{\varnothing\ne I\subseteq[n]}(-1)^{|I|+1}\frac{(n+1)!}{k+1}=\sum_{k=1}^n(-1)^{k+1}\binom{n}k\cdot\frac{(n+1)!}{k+1}\;.$$
But $\bigcup_{k\in[n]}P_k$ is just the set of permutations of $A$ in which $0$ does not come first, so
$$n!=(n+1)!-\sum_{k=1}^n(-1)^{k+1}\binom{n}k\cdot\frac{(n+1)!}{k+1}=(n+1)!\sum_{k=0}^n(-1)^k\binom{n}k\frac1{k+1}\;,$$
and dividing through by $(n+1)!$ yields the desired identity.
A: HINT:
$$\dfrac{\binom nk}{k+1}=\dfrac{n!}{k!\cdot(n-k)!\cdot(k+1)}=\dfrac1{n+1}\cdot\dfrac{(n+1)!}{\{n+1-(k+1)\}!\cdot(k+1)!}$$  $$=\dfrac1{n+1}\cdot\binom{n+1}{k+1}$$
Now $a=1,b=-1,m=n+1$ in
$$\sum_{k=0}^m\binom mk=(a+b)^m$$
