Density of order statistics I need help with order statistics:

Given a sample $X_1, \ldots, X_n$, $X_i \sim U_{0,1}$, i.e. the $X_i$ are uniformly distributed on $[0,1]$, determine the following for the corresponding order statistics:
a) the density of $X_{(k)}$
b) the joint density of $X_{(1)}, X_{(n)}$
c) the density of the range $R:=X_{(n)} - X_{(1)}$
d) the limit distribution for $2n(1-R)$ with $n \rightarrow \infty$.

Here is my idea for the first one:
a) For  the density of an order statistic we've shown:
$$f_{X_{(k)}}(t) = \binom{n}{k} k F_X(t)^{k-1}(1-F_X(t))^{n-k}f_X(t)$$
Given the fact that $X_i \sim U_{0,1}$, the density is pretty easy to determine, i.e.
$$f_{X_{(k)}}(t) = \binom{n}{k} k t^{k-1}(1-t)^{n-k} \mathbb{1}_{[0,1]}$$
b) For b), I think I can use the following formula:
$$f_{(i),(j)} = \dfrac{n!f(x_i)f(x_j)(F(X_i))^{i-1}(F(x_j)-F(x_i))^{j-1-i}(1-F(x_j))^{n-j}}{(i-1)!(j-1-i)!(n-j)!}$$ to get
$$f_{(1),(n)} (x_1,x_n) = (n-1)n(x_n-x_1)^{n-2}$$
Is that correct?
c) My idea was to use the transformation rule for densities, so
$$ \begin{pmatrix}
x \\
y 
\end{pmatrix} = \phi ( z,u) =  \begin{pmatrix}
z-u \\
u 
\end{pmatrix} $$
$$ \begin{pmatrix}
z \\
u 
\end{pmatrix} = \phi^{-1}(x,y) =  \begin{pmatrix}
x+u \\
y 
\end{pmatrix} $$ 
with $J_{\phi^{-1}}(x,y) = \begin{vmatrix}
1 & 0 \\
0 & 1 
\end{vmatrix} = 1$
Then $f_R = f_{(n)}(\phi^{-1}(x,y))f_{(1)}(\phi^{-1}(x,y))\cdot1 = \cdots$ - how do I proceed now?
d) Here I don't know how to start...
Thank you for the help!
 A: For b) as @air commented, independance is false. I would compute the joint CDF $F_{X_{(1)},X_{(n)}} (x,y) := \Bbb P [ X_{(1)} \leq x \cap X_{(n)} \leq y]$, and use the following lemma : 
If $F_{X,Y}$ is twice continuously differentiable,then $(X,Y)$ has density $f_{X,Y} = \frac {\partial ^2f}{\partial x \partial y}F_{X,Y}$. (The result holds with weaker conditions, using the more general form of the differenciation under integral theorem)
The rest follows, your idea of using change of variables is good.
A: \begin{align}
f_{X_{(k)}} (x) & = \frac d {dx} \Pr( X_{(k)} \le x) \\[10pt]
& = \frac d {dx} \Pr\left( \text{at least $k$ of $X_1,\ldots,X_n$ are} \le x \right) \\[10pt]
& = \frac d {dx} \sum_{j=k}^n \binom n j x^j (1-x)^{n-j}.
\end{align}
Now we come to the part that poses the principal difficulty in finding the answer. Differentiating each term in this sum relies on the product rule, except in the term where $j=n$. Here's what happens
\begin{align}
& \frac d {dx} \left( \cdots + \binom n \ell x^\ell(1-x)^{n-\ell} + \binom n {\ell+1} x^{\ell+1} (1-x)^{n-\ell-1} + \cdots \right) \\[15pt]
= {} & \cdots +{} \binom n \ell \left(\ell x^{\ell-1} (1-x)^{n-\ell} - x^\ell (n-\ell)(1-x)^{n-\ell-1}\right) \tag a \\[10pt]
& \phantom{\cdots+{}} \binom n {\ell + 1} \left( (\ell+1)x^\ell(1-x)^{n-\ell-1} - x^{\ell+1} (n-\ell-1) (1-x)^{n-\ell-2}  \right) + \cdots \tag b
\end{align}
Observe that the second term in line (a) cancels out the first term in line (b) because
$$
-\binom n \ell (n-\ell) + \binom n {\ell+1} (\ell+1) = 0.
$$
The effect is that all terms except the first one --- the one in which $j=k$ --- cancel out.
