Proving complex trigonometric identity using power series 
Prove $2$cos$^2(z) = 1+$cos$(2z)$ using power series.

I know that cos$(z) = \sum (-1)^n\frac{z^{2n}}{(2n)!}$
I also know that if 
$a(z) = \sum a_nz^n$ and $b(z) = \sum b_nz^n$
then 
$a(z)b(z) = \sum c_nz^n $ 
where 
$c_n = a_0b_n + a_1b_{n-1}+a_2b_{n-2}+...+a_nb_{0}$
Combining this I got
cos$^2(z) = \sum \left [   (-1)^nz^{2n} \sum_{k=0}^{n} \frac{1}{(2n-2k)!(2k)!} \right]$
and
cos$(2z) = \sum \left [(-1)^nz^{2n} \frac{4^n}{(2n)!} \right]$ 
I havent really gotten further than this.
I tried writing $\sum_{k=0}^{n}\frac{1}{(2n-2k)!(2k)!}$ 
as 
$\frac{1}{(2n)!}\sum_{k=0}^{n}\binom{2n}{2k}$
but no luck!
Any ideas?

EDIT

So we have cos$^2(z) = \sum \left [   (-1)^nz^{2n} \sum_{k=0}^{n} \frac{1}{(2n-2k)!(2k)!} \right]$
$= \sum \left [  \frac{ (-1)^n}{(2n)!}z^{2n} \sum_{k=0}^{n} \binom{2n}{2k}  \right]$ 
$= \sum \left [  \frac{ (-1)^n}{(2n)!}z^{2n} \frac{4^n}{2}  \right]$ 
$= \frac{1}{2} \sum \left [ (-1)^n \frac{4^n}{(2n)!}z^{2n}  \right]$ 
$= \frac{1}{2}$cos$(2z)$
Therefore $2$cos$^2(z) = $cos$(2z)$
 A: The $n$th term of $\cos^2(z)$ is $$\sum_{k=0}^n (-1)^n \frac{z^{2n}}{(2k)!(2n-2k)!}  $$
By multiplying and dividing by $(2n)!$ we get $\frac{1}{(2n)!}\sum_{k=0}^{n} \frac{(2n)!}{(2n-2k)!2k!}=\frac{1}{(2n)!}\sum_{k=0}^n{2n\choose 2k}$.
Now $\sum_{k=0}^n{2n\choose 2k}$ is a very curious number for $n>0$ and is equal to $\frac{4^{n}}{2}$ because of the following two lines:
$$0=(1-1)^{2n}=\sum_{k=0}^{2n}(-1)^k{2n \choose k}=\sum_{k=0}^{n}{2n \choose 2k} -\sum_{k=0}^{n-1}{2n \choose 2k+1}.$$
But
$$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}{2n \choose k}=\sum_{k=0}^{n}{2n \choose 2k} +\sum_{k=0}^{n-1}{2n \choose 2k+1}.$$
So our lovely sum $ \sum_{k=0}^{2n}{2n \choose k}$ is exactly half of $4^n$. Now it must be evident to you why we need $2\cos^2(z)$ in our equation. I am confident that you can finish the argument from here after understanding how this multiplying by $2$ affects the pesky degree $0$ term.
A: I needed to use the "higher-derivatives" form of the Product Rule in a problem I was showing to students, which put me in mind of this approach.
While it sounds tedious to construct the Maclaurin series for $ \ \cos^2 z \ $ by taking successive derivatives, the properties of the cosine function save us some trouble.  We will label $ \ f(z) \ = \ \cos z \ $ and call the function in question $ \ g(z) \ = \ [ \ f(z) \ ]^2 \ $ .  We can use the facts that $ \ f \ ' \ = \ -f''' \  = \ 0 \ $ and $ \ f \ = \ f'' \ = \ 1 \ $ at $ \ z \ = \ 0 \ $ .  At this point, then, the successive derivatives are
$  g(0) \ = \ f^2(0) \ = \ 1 \ $ ; 
$  g'(0) \ = \ 2  \ f \ ' \ f \ (0) \ = \ 0 \ $ ;
$ g''(0) \ = \ 2 \ [ \ f'' \ f \ + \ (f \ ')^2 \  ] (0) \ = \  2  \ \cdot \ (-1)  \ \cdot \ 1 \ + \ 0 \ = \ -2 \ $ ;
$ g'''(0) \ = \ 2 \ [ \ f''' \ f \ + \ 2 \ f'' \ f \ ' \ + \ f \ ' \ f'' \ ] (0) \ = \ 0 \ $
(since every term contains an odd derivative of cosine);
[it begins to emerge, if one is unfamiliar with this relation, that this extension of the Product Rule has a form reminiscent of the Binomial Theorem]
$ g^{(4)}(0) \ = \ 2 \  [ \ f^{(4)} \ f \ + \ 3 \ f''' \ f \ ' \ + \ 3 \ (f'')^2 \ + \ f \ ' \ f''' \ ] (0) \ = \ 2 \ ( \ 1 \ \cdot \ 1 \ + \ 3 \ \cdot \ [-1]^2 \ ) \ = \ 8 \ $ ;
[it is also clear by this point that the odd derivatives of $ \ g \ $ will comprise terms containing only odd derivatives of $ \ f \ $ , thus they are all equal to zero]
$ g^{(6)}(0) \ = \ 2 \  [ \ f^{(6)} \ f \ + \ 5 \ f^{(5)} \ f \ ' \ + \ 10 \ f^{(4)} \ f'' \ + \ 10 \ (f''')^2 \ + \ 5 \ f'' \ f^{(4)} \ + \ f \ ' \ f^{(5)} \ ] (0) $
$ = \ 2 \ ( \ [-1] \ \cdot \ 1 \ + \ 10 \ \cdot \ 1 \ \cdot \ [-1] \  + \ 5 \ \cdot \ [-1] \ \cdot \ -1 \ \  ) \ = \ -32 \ $ . 
We see now two features of these sums.  The first is that the non-zero terms for the derivatives $ \ g^{(2m)} \ $  produce only positive products from the powers of $ \ 1 \ $ and $ \ (-1) \ $, while the derivatives  $ \ g^{(2m+2)} \ $ have only products that are negative.  The second is that the non-zero terms in the sums alternate, so that "doubling" term gives us in effect a "complete row of binomial coefficents" (corresponding to the "odd" rows of "Pascal's Triangle") .  We can then use a familiar identity for these coefficients , $ \ \sum_{j = 0}^n \ \binom{n}{j} \ = \ 2^n \ $ , to conclude that
$$ g^{(2k)} \ \ =  \ \ (-1)^k \ \cdot \ 2^{2k-1} \ \ . $$
The Maclaurin series for $ \ \cos^2 z \ $ is thus $$ \ 1 \ + \ \sum_{n=\mathbf{1}}^{\infty} \ \frac{(-1)^n \ \cdot \ 2^{2n-1}}{(2n) \ !} \ z^{2n} \ \ , $$
the first term not being compatible with the general term of the series.
The series for $ \ \cos(2z) \ $ can be constructed as
$$ \sum_{n=0}^{\infty} \ \frac{(-1)^n }{(2n) \ !} \ (2z)^{2n} \ \ = \ \ \sum_{n=0}^{\infty} \ \frac{(-1)^n \ \cdot \ 2^{2n} }{(2n) \ !} \ z^{2n} \ \  ; $$
the first term of which may be extracted to write
$$   \cos(2z) \ \ = \ \  \frac{(-1)^0 \ \cdot \ 2^0 }{0 \ !} \ + \ 2 \ \sum_{n=\mathbf{1}}^{\infty} \ \frac{(-1)^n \ \cdot \ 2^{2n-1} }{(2n) \ !} \ z^{2n} $$ $$ = \ \ 1   \ + \ 2 \ \sum_{n=\mathbf{1}}^{\infty} \ \frac{(-1)^n \ \cdot \ 2^{2n-1} }{(2n) \ !} \ z^{2n} \ \ . $$
Lastly, we can use our earlier series to obtain
$$   \cos(2z) \ \ = \ \   1   \ + \ 2 \ (\cos^2 z \ - \ 1) \ \ = \ \ 2 \ \cos^2 z \ - \ 1 \ \ , $$
thereby establishing our identity.
