Why is substitution allowed in Taylor Series? While finding the Taylor Series of a function, when are you allowed to substitute? And why?
For example:
Around $x=0$ for $e^{2x}$ I apparently am allowed to substitute $u=2x$ and then use the known series for $e^u$. But for $e^{x+1}$ I am not allowed to substitute $u=x+1$.
I know the technique for finding the Taylor Series of $e^{x+1}$ around $x=0$ by taking $e^{x+1}=e\times e^x$. However, I am looking for understanding and intuition for when and why it is allowed to apply substitution.
Note: there are several question that are similar to this one, but I have found none that actually answers the question "why"; or that shows a complete proof.

EDIT: Thanks to the answer of Markus Scheuer I should refine the question to cases where the series is finite, for example $n\to3$
 A: 
A function $f(x)$ analytic at $x=0$ can be represented as power series within an open disc with radius of convergence $R$.
  \begin{align*}
f(x)=\sum_{n=0}^\infty a_nx^n\qquad\qquad \qquad |x|<R
\end{align*}
Any substitution $x=g(u)$ is admissible as long as we respect the radius of convergence.
  \begin{align*}
f(g(u))=\sum_{n=0}^\infty a_n \left(g(u)\right)^n\qquad\qquad\quad |g(u)|<R
\end{align*}

We know $f(u)=e^u$ can be represented as Taylor series convergent for all $u\in\mathbb{R}$, i.e. the radius of convergence $R=\infty$.
\begin{align*}
f(u)=e^u=\sum_{n=0}^\infty \frac{u^n}{n!}\qquad\qquad\qquad u\in \mathbb{R}
\end{align*}

Substitution $u=2x$
We consider
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad 2x\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$2x\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(2x)=e^{2x}$ is $\infty$.

We obtain
\begin{align*}
f(2x)=e^{2x}=\sum_{n=0}^\infty \frac{(2x)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}

Substitution $u=x+1$
We consider
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x+1\in \mathbb{R}
\end{align*}
This substitution is admissible for all $x \in \mathbb{R}$ since $$x+1\in\mathbb{R}\qquad\Longleftrightarrow\qquad x\in\mathbb{R}$$ So, the radius of convergence of the Taylor series of $f(x+1)=e^{x+1}$ is $\infty$.

We obtain
\begin{align*}
f(x+1)=e^{x+1}=\sum_{n=0}^\infty \frac{(x+1)^n}{n!}\qquad\qquad\qquad x\in \mathbb{R}
\end{align*}

We also obtain
  \begin{align*}
e\cdot e^x&=\left(\sum_{k=0}^\infty \frac{1}{k!}\right)\left(\sum_{l=0}^\infty \frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{{k+l=n}\atop{k,l\geq 0}}\frac{1}{k!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty \left(\sum_{l=0}^n\frac{1}{(n-l)!}\cdot\frac{x^l}{l!}\right)\\
&=\sum_{n=0}^\infty\left(\sum_{l=0}^n\binom{n}{l}x^l\right)\frac{1}{n!}\\
&=\sum_{n=0}^\infty\frac{(x+1)^n}{n!}\\
&=e^{x+1}
\end{align*}

Conclusion: We can use any substitution for convenience as long as we respect the radius of convergence.
A: If $ f(x)=P_n(x)+x^n\epsilon(x)$ then
$f(u(x))=P(u(x))+(u(x))^n\epsilon(u(x))$
with $\lim_{x\to 0}\epsilon(x)=0$.
thus we need that
$\lim_{x\to 0}\epsilon(u(x))=0$
so, we must have
$$\lim_{x\to 0}u(x)=0$$
A: The quantity $2x$ is a product and as $x\to 0$ it remains a small number.
The quantity $x+n$ for $n\neq 0$ is not a little quantity anymore, and so you are not anymore around zero but you're around $n$.
