How did this result come about? I was reading Chebyshev polynomials Wiki page and I could not understand one thing
$$ T_n(x)  = x^n \sum_{k=0}^{\left  \lfloor \frac{n}{2} \right \rfloor} \binom{n}{2k} \left (1 - x^{-2} \right )^k \\$$
$$= \tfrac{n}{2} \sum_{k=0}^{\left \lfloor \frac{n}{2} \right \rfloor} (-1)^k \frac{(n-k-1)!}{k!(n-2k)!}~(2x)^{n-2k}  \\ $$
How did this step came about? They probably are missing one step in between. As far as I know, on needs to expand $\left (1 - x^{-2} \right )^k$ and then change the order of summation but that may take quite a time and it is not this easy, I guess. 
 A: Suppose we have the Chebychev polynomial
$$T_n(x) = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \left(1-\frac{1}{x^2}\right)^k
\\ = x^n \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} \frac{1}{x^{2k}} 
\sum_{q=0}^k {k\choose q} (-1)^{k-q} x^{2q}$$
and we seek to verify that
$$[x^{n-2p}] T_n(x)
= \frac{1}{2} n (-1)^p \frac{1}{n-p} {n-p\choose p} 2^{n-2p}.$$
For this coefficient we must have $2q-2k = -2p$
or $q = k - p$
getting
$$ \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose k-p} (-1)^p
\\ = \sum_{k=0}^{\lfloor n/2 \rfloor}
{n\choose 2k} {k\choose p} (-1)^p.$$
To evaluate this introduce for the sum term
$${n\choose 2k} = {n\choose n-2k} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2k+1}} (1+z)^n \; dz.$$
This is zero when $2k\gt n$ so we may extend $k$ to infinity,
getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n 
\sum_{k\ge 0} {k\choose p} (-1)^p z^{2k} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n 
\sum_{k\ge p} {k\choose p}  z^{2k} \; dz 
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n+1}} (1+z)^n 
\sum_{k\ge 0} {k+p\choose p} z^{2k+2p} \; dz 
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2p+1}} (1+z)^n 
\frac{1}{(1-z^2)^{p+1}} \; dz
\\ = (-1)^p \frac{1}{2\pi i}
\int_{|z|=\epsilon}
\frac{1}{z^{n-2p+1}} (1+z)^{n-p-1} 
\frac{1}{(1-z)^{p+1}} \; dz.$$
Exract the coefficient to get
$$(-1)^p \sum_{q=0}^{n-2p} 
{n-p-1\choose q} {n-2p-q+p\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p} 
{n-p-1\choose n-p-q-1} {n-p-q\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p} \frac{n-p-q}{n-p}
{n-p\choose n-p-q} {n-p-q\choose p}
\\ = (-1)^p \sum_{q=0}^{n-2p} \frac{n-p-q}{n-p}
{n-p\choose p} {n-2p\choose q}
\\= (-1)^p {n-p\choose p} \frac{1}{n-p}
\sum_{q=0}^{n-2p} (n-p-q)
{n-2p\choose q}.$$
Hence it remains to show that
$$\frac{1}{2} n 2^{n-2p} =
\sum_{q=0}^{n-2p} (n-p-q) {n-2p\choose q}.$$
This is
$$(n-p) 2^{n-2p} 
- \sum_{q=0}^{n-2p} q {n-2p\choose q}
= (n-p) 2^{n-2p} 
- \sum_{q=1}^{n-2p} q {n-2p\choose q}
\\ = (n-p) 2^{n-2p} 
- (n-2p) \sum_{q=1}^{n-2p} {n-2p-1\choose q-1}
\\ = (n-p) 2^{n-2p}  - (n-2p) 2^{n-2p-1}
= n 2^{n-2p} - n 2^{n-2p-1} = \frac{1}{2} n 2^{n-2p}.$$
This concludes the argument.
