Cauchy product of two different Taylor series $e^x$ and $\cos x$ I have to do the following task that goes like this: 
Calculate the first 5 terms of Taylor expansion of function $f(x) = e^x \cos x$ centered in point $x_0 = 0$ by:
a) forming Cauchy product of Taylor series from $e^x$ and $\cos x$ at point $0$,
b) by finding the derivatives of $f(x)$ and inserting them into the general formula for the Taylor series.
I will now write how I think the first part of the assignment should be done, please correct me if I'm wrong.
Taylor series formula: 
$$\sum_{n=0}^{\infty}{\frac{f^{(n)}(x_0)}{n!} (x-x_0)^n}$$
First 5 terms when $x_0 = 0$:
$$\begin{align}
&f(x) = e^x\cos x = 1\\
&f'(x) = e^x(\cos x - \sin x) = 1\\
&f''(x) = e^x(-2\sin x) = 0\\
&f'''(x) = 2e^x(-\sin x - \cos x) = -2\\
&f''''(x) = 2e^x(-2\cos x) = -4
\end{align}
$$
a) We can represent $e^x$ and $\cos x$ as Taylor series:
$$e^x = \sum_{n=0}^{\infty}{\frac{x^n}{n!}} \quad\text{and}\quad\cos x = \sum_{n=0}^{\infty}{(-1)^n \frac{x^{2n}}{(2n)!}}$$
Now, we also have formula for Cauchy product, which is: $$(\sum_{i=0}^{\infty}{a_i}) (\sum_{j=0}^{\infty}{b_j}) = \sum_{k=0}^{\infty}{c_k}$$, which gives us: $$c_k = \sum_{k=0}^{n}{a_kb_{n-k}}$$
Now I'm starting to be confused, since I couldn't find anywhere solution for the Cauchy product of these to series, so I just tried this (again, please correct me if it's wrong):
$$\sum_{k=0}^{n}{\frac{x^k}{k!}(-1)^{n-k}\frac{x^{2n-2k}}{(2n-2k)!}} = \sum_{k=0}^{n}{(-1)^{n-k}\frac{x^{2n-k}}{k!(2n-2k)!}}$$
If this is correct, is this the final solution? Also, why it was said "at point $0$", should I substitute something with $0$?
b) I don't have a clue what are we suppose to do here. Please help :)
Lastly, we are asked to establish the principle (general formula?) of $n$-th derivative ($n \in \Bbb{N}$) of $f(x)$ and give the complete Taylor expansion of $f(x)$ in point $0$.
P.S. I was translating from German to English, which both aren't my mother languages, so maybe I have mistranslated something.
Sorry for the long post.
Thank you in advance
 A: Hint for (a). Note that $a_n=1/n!$ and $b_{2n}=(-1)^n/(2n)!$ (and zero otherwise). Hence
$$\frac{f^{(n)}(0)}{n!}=c_n = \sum_{k=0}^{n}{a_kb_{n-k}}=\sum_{k=0}^{n}a_{n-k}b_{k}=\sum_{k=0}^{\lfloor n/2\rfloor}a_{n-2k}b_{2k}=\sum_{k=0}^{\lfloor n/2\rfloor}\frac{1}{(n-2k)!}\cdot \frac{(-1)^k}{(2k)!}.$$
Hint for (b). By the general Leibniz rule we have that
$$f^{(n)}(0)=\sum_{k=0}^n \binom{n}{k} D^{(n-k)}(e^x)_{x=0}\cdot D^{(k)}(\cos(x))_{x=0}\\=\sum_{k=0}^n \binom{n}{k} \cdot D^{(k)}(\cos(x))_{x=0}
=\sum_{k=0}^n \binom{n}{2k} (-1)^k.$$
By the way note that  $f(x)=\mbox{Re}(e^{x(1+i)})$ and therefore 
$$f^{(n)}(0)=\mbox{Re}( (1+i)^n).$$
A: Since at first all you're asked for are the first five terms you don't need the fancy formula for the Cauchy product. Just start multiplying and collecting everything up to degree four:
$$
e^x \cos x = \left( 1 + x + \frac{x^2}{2}  + \cdots \right)
\left( 1 - \frac{x^3}{6}  + \cdots \right) .
$$
Then you add to your question 

Lastly, we are asked to establish the principle (general formula?) of
  $n$-th derivative  of $f(x)$ and give the complete Taylor expansion of
  $f(x)$ in point $0$.

I think that requires a formula like the one in @RobertZ 's answer.
