Evaluate $\int_{0}^{1}(1-x)^ndx$ by expanding the bracket. I'd like to get a hint on this exercise. I believe I'm somewhat close to the answer. I used the binomial theorem to get:
$\displaystyle\int_{0}^{1}(1-x)^ndx = \int_{0}^{1}\sum_{k=0}^{n}{n\choose k}(-x)^kdx = \sum_{k=0}^{n}\left\{\binom{n}{k}(-1)^k\int_{0}^{1}x^kdx\right\} = \sum_{k=0}^{n}\binom{n}{k}(-1)^k\frac{x^{k+1}}{k+1}\bigg|_0^1 = \sum_{k=0}^{n}\frac{(-1)^kn!}{(k+1)!(n-k)!}$ 
I've proved that this is equal to $\frac{1}{n+1}$ (which by substitution I found to be the answer) for n even by writing down the sum for some terms  and checking that they cancel eachother except for k = n. I found that I can't cancel the terms for n odd, and so I tried to prove by induction that $\sum_{k=0}^{n}\frac{(-1)^kn!}{(k+1)!(n-k)!} = \frac{1}{n+1} \forall n$ and got nowhere.
 A: Let $$\eqalign{F(t) &= \sum_{k=0}^n \dfrac{n!}{(k+1)!(n-k)!} t^{k+1}\cr
             G(t) &= \dfrac{(t+1)^{n+1}-1}{n+1} }$$
I claim $F(t) = G(t) $ for all $t$.
At $t = 0$, both sides are equal to $0$.  Now
$$ F'(t) = \sum_{k=0}^n \dfrac{n!}{k!(n-k)!} t^k = (t+1)^n$$
(by the binomial theorem), and of course this is $G'(t)$.
We conclude that $F(t) = G(t)$.  In particular, take $t=-1$.
A: Using the binomial theorem, we find
$$
(n+1)\sum_{k=0}^n {(-1)^k n! \over (k+1)!(n-k)!} = \sum_{k=0}^n \frac{(-1)^k (n+1)!}{(k+1)!(n-k)!} = \sum_{k=0}^n (-1)^k \binom{n+1}{k+1} \\ = -\sum_{\ell=1}^{n+1} (-1)^{\ell} \binom{n+1}{\ell} = - \left( (1-1)^{n+1} - \binom{n+1}{0} \right) = 1.
$$
(Of course, to prove that $\int_0^1 (1-x)^n dx = \frac{1}{n+1}$, one could also note that writing $x=1-y$, we have $\int_0^1 (1-x)^n dx = - \int_1^0 y^n dy = \int_0^1 y^n dy = \frac{1}{n+1}$.)
A: Hint: you have the following binomial coefficient identity:
$$ \frac{n+1}{k+1} \binom {n}{k} = \binom{n+1}{k+1},  $$
which you can prove by fiddling about with the factorials.
A: Compute: $$(n+1) \cdot \sum_{k=0}^n {n \choose k} (-1)^k \frac{1}{k+1}$$
$$\frac{n!}{k! (n-k)!} \cdot \frac{n+1}{k+1} = \frac{(n+1)!}{(k+1)! (n-k)!}$$
Note that $(n-k) = ((n+1) - (k+1))$ thus $$\frac{n+1}{k+1} {n\choose k} = {n+1 \choose k+1 }$$
Therefore the sum becomes: $$-\sum_{k=0}^n (-1)^{k+1} {n+1 \choose k+1}$$
When $n$ is odd, subtract and add $${n+1 \choose 0}=1$$, reindexing the sum gives:
$$\left(-\sum_{k=0}^{n+1} (-1)^{k} {n+1 \choose k}\right) + 1 = -(1-1)^{n+1}+1=1$$
