Taylor Series of $e^{x^{2}+x}$ about $0$ and about $1$ Find the taylor series of $e^{x^{2}+2x}$ about $0$. 
I would normally take $e^{x}=\sum^{\infty}_{n=0}\frac{x^n}{n!}$ and then plug in $x^{2}+2x$, so: $\sum^{\infty}_{n=0}\frac{(x^2+2x)^n}{n!}$, but I am not sure under what conditions we can use this "plugging in" for taylor polynomials?
My guess would be that the taylor series simply needs to be around $0$?(Correct me if I am wrong). If this is so, how would I compute $e^{x^{2}+2x}$ about $1$?
Idea: I take $y=x-1$, so:
$f(1+y)=e^{(1+y)^{2}+2(1+y)}= \sum_{0}^{\infty}\frac{((1+y)^{2}+2(1+y))^{n}}{n!}$ and then I do not know what to do. 
Any remarks on when I can use the shortcut for "plugging in" taylor series would be greatly appreciated
 A: We can find a  series  expansion of $e^{x^2+2x}$ for instance  via Cauchy-multiplication of  series.

Expanding at $x=0$ we obtain
  \begin{align*}
\color{blue}{e^{x^2+2x}}&=e^{x^2}e^{2x}\\
&=\left(\sum_{k=0}^\infty\frac{(x^2)^k}{k!}\right)\left(\sum_{l=0}^\infty\frac{(2x)^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{{2k+l=n}\atop{k,l\geq 0}}\frac{2^l}{k!l!}\right)x^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(\sum_{k=0}^{\lfloor n/2 \rfloor}\frac{n!}{k!(n-2k)!}2^{n-2k}\right)\frac{x^n}{n!}}
\end{align*}

Similarly we obtain an expansion at $x=1$ using one Cauchy series multiplication via
\begin{align*}
e^{x^2+2x}=e^{(x-1)^2}e^{x-1}e^{-3}
\end{align*}
A: You can plug in $x^2+2x$ into the Taylor series for $e^x$:
\begin{align}
e^{x^2+2x} &= \sum_{n=0}^\infty \frac{(x^2 + 2x)^n}{n!}\\
&= \sum_{n=0}^\infty \frac1{n!}\sum_{k=0}^n {n \choose k} x^{2k}2^{n-k}x^{n-k}\\
&= \sum_{n=0}^\infty \left[\frac1{n!}\sum_{k=0}^n {n \choose k} 2^{n-k}\right]x^{n+k}\\
&= \text{set } [m = n+k]\\
&= \sum_{m=0}^\infty \left[\sum_{n=0}^\infty \left[\frac1{n!}\sum_{\substack{k=0\\n+k = m}}^n {n \choose k} 2^{n-k}\right]\right]x^{m}\\
&= \sum_{m=0}^\infty \left[\sum_{n=0}^m\frac1{n!}{n \choose m-n}2^{2n-m}\right]x^{m}\\
\end{align}
Or you can use the Cauchy product:
\begin{align}
e^{x^2+2x} &= e^{x^2}e^{2x}\\
&= \left(\sum_{n=0}^\infty \frac{x^{2n}}{n!}\right)\left(\sum_{k=0}^\infty \frac{2^{k}x^k}{k!}\right)\\
&= \sum_{n=0}^\infty \left(\sum_\limits{\substack{k=0\\k \text{ even }}}^n\frac{1}{(k/2)!}\frac{2^{n-k}}{(n-k)!}\right)x^n\\
\end{align}
