Is there anything wrong with my method below? Also, is there an easier method?
For $sin\,x = \sum^{\infty}_{k=0}\frac{(-1)^kx^{1+2k}}{(1+2k)!}$,
$L_1 = \lim_{k\rightarrow\infty} \left| \frac{x^{1+2(k+1)}}{(1+2(k+1))!} \div \frac{x^{1+2k}}{(1+2k)!}\right|$
$= \lim_{k\rightarrow\infty} \left| \frac{x^2}{(2k+3)(2k+2)} \right| = 0 < 1 \;\forall x \in \mathbb{R} $
Thus, by ratio test, $sin\,x = \sum^{\infty}_{k=0}\frac{(-1)^kx^{1+2k}}{(1+2k)!}$ converges $\;\forall x \in \mathbb{R} $
Next, for $e^{sin\,x}= 1 + \sum^\infty_{k=1} \frac{sin^k x}{k!}$,
$L_2 = \lim_{k\rightarrow\infty} \left| \frac{sin^{k+1}x}{(k+1)!} \div \frac{sin^k x}{k!}\right|$
$= \lim_{k\rightarrow\infty} \left| \frac{sin\,x}{k+1} \right|$
$= \lim_{k\rightarrow\infty} \left| \frac{\sum^{\infty}_{k=0}\frac{(-1)^kx^{1+2k}}{(1+2k)!}}{k+1} \right| = 0 < 1 \;\forall x \in \mathbb{R}$
since $\sum^{\infty}_{k=0}\frac{(-1)^kx^{1+2k}}{(1+2k)!}$ converges $\forall x \in \mathbb{R}$
Thus, by ratio test, the Taylor's series (about 0) of $e^{sin\,x}$ converges $\forall x \in \mathbb{R}$
I don't think there's anything wrong with what you've written.
I wouldn't say it was necessary to sub back in the taylor series for $\sin{x}$ three lines from the bottom. I'm not sure you need to use the taylor series for $\sin$ explicitly anyway. You could either note that
$\forall x \in \mathbb{R},\:\mid \sin{x}\mid\; \leq 1$
and then put that into your limit and have $L_2\leq0$, hence $L_2=0$.
Or just use that sin x is constant w.r.t. the limit being taken and take it outside the limit.
$L_2= \lim\limits_{k\rightarrow\infty} \left| \frac{\sin x}{k+1} \right|=\left| \sin x \right|\lim\limits_{k\rightarrow\infty}\frac{1}{k+1}=0$.
I suppose this holds whenever $\sin$ is well defined, which is everywhere.