McLaurin series expansion from known series The goal is to find the Mclaurin series from a known series.
$$f(x) = x^3e^{x^2-2} $$ and evaluate  $ f^{2015}(0)$
My solution:
1.)
$$f(x) = e^x $$
$$e^x = \sum_{n=0} ^\infty = \frac{x^n}{n!} $$
$$e^{x^2} = \frac{x^{2n}}{n!}$$ $$e^{x^2 - 2} = e^{x^2}.e^{-2} $$ 
$$\frac{e^{x^2}}{e^2} [*]$$ $$x^3e^{x^2-2} = \sum_{n=0}^ \infty \frac{x^{2n+3}}{e^2n!}$$
and on applying the ratio test to check for its convergence I arrive at $0$ and it does converge!
2.) $$f^{2015}(0) = 2n+3 = 2015 $$
and $ n = 1006 \in \Bbb N$ and $ \frac{2015!}{e^2.1006!} $
I'm not really sure about the answers for both the parts. Are they written legally? Have I made mistakes algebraically?
* I'm not sure if it has to be written that way or to be included in the numerator.
 A: *

*This is basically correct, but I think that you can tighten it up a bit, and/or clean up the formatting.  I might have responded as follows:

\begin{align}
f(x)
 &= x^3 e^{x^2-2} \\
 &= x^3 e^{-2} e^{x^2} \tag{$\ast$} \\
 &= x^3 e^{-2} \sum_{n=0}^{\infty} \frac{(x^2)^n}{n!} \tag{1} \\
 &= x^3 e^{-2} \sum_{n=0}^{\infty} \frac{x^{2n}}{n!} \\
 &= e^{-2} \sum_{n=0}^{\infty} \frac{1}{n!} x^{2n+3} \tag{$\ast$} \\
 &= e^{-2} \sum_{n=0}^{\infty} \frac{1}{n!} \frac{(2n+3)!}{(2n+3)!} x^{2n+3} \tag{$\ast$} \\
 &= \sum_{n=0}^{\infty} \left[ \frac{(2n+3)!}{e^2 n!} \right] \frac{1}{(2n+3)!} x^{2n+3} \tag{2} \\
 &= \sum_{k=0}^{\infty} C_k \frac{1}{k!} x^k,
\end{align}
  Where
  $$ C_k = f^{(k)}(0) = \begin{cases}
\dfrac{(2n+3)!}{e^2 n!} & \text{if $k=2n+3$ for some $n\in\mathbb{N}$, and} \\
0 & \text{otherwise}.
\end{cases}$$
  At (1), we expand $e^{x^2}$ using the MacLaurin series for $e^t$ with $t=x^2$.  Note that $x^3e^{-2}$ is constant with respect to this series, and so the convergence of (2) follows from the convergence of the MacLaurins series for $e^t$.

The ($\ast$)s indicate steps that are, perhaps, overly computationally pedantic.  If I were grading this, and a student left out one or more of those steps, I'd probably still give the work full marks.  That being said, I don't think it hurts to be pedantic when you are doing work as part of an assignment, rather than for publication.
Note, also, the conclusion.  The series at the end is written as a power series, indexed over $\mathbb{N}$.  From this, it is much easier to see what the $k$-th term (and therefore the $k$-th derivative) is meant to look like, which makes it much easier to justify your answer in (2).
Finally, observe that we don't actually need to invoke the ratio test to show that the MacLaurin series at the end converges (we can, but we don't have to).  Careful bookkeeping throughout does the job for us.

*If a student handed in the work you have provided for this problem, I would probably dock them a few points, since it is not clear how you got from your first answer to the equation
$$ f^{(2015)}(0) = 2n+3 = 2015. $$
Indeed, this isn't even correct, so far as I can tell (though I think I understand what you are trying to say).  I might provide the answer

From part (1), we know that
  $$ f^{(2015)}(0) = C_{2015}, $$
  where the value of $C_{2015}$ depends on whether or not $2015 = 2n+3$ for some natural number $n$.  Solving, we obtain
  $$ 2015 = 2n + 3 \implies 2n = 2012 \implies n = 1006. $$
  Hence we can write $2015 = 2n+3$ for the natural number $n = 1006$, and so
  $$ f^{(2015)}(0) = C_{2015} = \frac{2015!}{e^2 1006!}.$$

