Show that if $A$ and $B$ commute, then $B$ commutes with $e^A$ Show that if $A$ and $B$ commute, then $B$ commutes with $e^A$
For the first one I have $$e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} +\dots$$
I believe I can pull out $$I + A + \frac{A^2}{2!} + \frac{A^3}{3!} +\dots$$ and multiply by $B$ on either side but not sure the validity of this proving this
 A: Your approach to the problem is nice. Expanding the exponent $e^A$ is a key step in proving that if $A$ and $B$ commute then $e^A$ and $B$ also commute.
It is given to us that $A$ and $B$ commute and thus we can write: $$\left[A,B\right] = \left[B,A\right]$$ Note:If $A$ and $B$ are matrices this also means that $A$ and $B$ are both square matrices.
Now, consider an arbitrary exponent of $A$ i.e. $A^k$. Let us multiply this by $B$: $$A^kB = (AAA\cdots k\text{ times}) B$$ Now, since $A$ and $B$ commute, we can interchange the position of the last $A$ in the RHS of the previous product and $B$. $$\implies A^kB = (AAA\cdots k-1\text{ times}) BA$$ This can be done repeatedly and $B$ can be moved to an arbitrary position within the product: $$\implies A^kB = (AAA\cdots k-n\text{ times})B(AAA\cdots n\text{ times})$$ In particular we can place $B$ in the beginning of the product:$$\implies A^kB = B(AAA\cdots k\text{ times})$$ Now, Let us prove that $e^A$ and $B$ commute. As you correctly mentioned:
$$
\begin{align}
&\boxed{e^A = \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}}\\ 
\implies \left[ e^A,B\right] = e^AB &= \left[ \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}\right ]B\\
\implies e^AB &= \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}B\\
&= \Sigma_{i=0}^{\infty}\dfrac{A^iB}{i!}\\
&=\Sigma_{i=0}^{\infty}\dfrac{BA^i}{i!}\\
&=\Sigma_{i=0}^{\infty}B\dfrac{A^i}{i!}\\
&=B \left[ \Sigma_{i=0}^{\infty}\dfrac{A^i}{i!}\right] = \left[ B,e^A\right] 
\end{align}\\
$$
Hence proved that $e^A$ and $B$ also commute.
A: Possibly overkill, but a technique that is worth knowing:
Define $U(x)=[B, e^{x A}]$, then ${d U \over dx} = [B, Ae^{x A}]$. Since $[A,B]=0$, we have $${d U \over dx} = BAe^{x A} - A e^{x A}B = A[B, e^{x A}] = A U(x)$$
Solving this: $$U(x)=e^{x A} U(0)$$  and since $$U(0)=[B, e^{0A}]=[B,I]=0$$ we deduce that $$[B, e^A]=U(1)=e^{A}U(0)=0$$
