Let's look at the series first.
$$\sin(x)=\sum \limits_{k=0}^{\infty} (-1)^{k} \frac{x^{2k+1}}{(2k+1)!}$$
For each time you differentiate, you need to substract one from the exponent of $x$, i.e.
$$\frac{\mathrm{d}x^k}{\mathrm{d}x}=k\cdot x^{k-1}$$
However note, that we can not abbreviate sequential differentiation in general as
$$\frac{\mathrm{d^n}x^k}{\mathrm{d}x^n}=\frac{k!}{(k-n)!}x^{k-n}$$
An easy counter example would be
$$\frac{\mathrm{d^3}x^2}{\mathrm{d}x^3}=2\frac{\mathrm{d^2}x}{\mathrm{d}x^2}=2\frac{\mathrm{d}}{\mathrm{d}x}1 = 0$$
which is obviously different from $x^{2-3}=x^{-1}$. So this only holds if $n \leq k$.
To prevent this from happening in our infinite series, we need to shift the index accordingly, so we don't get these terms, that would end up zero in between.
$$\frac{\mathrm{d}^2}{\mathrm{d}x^2}\sin(x)=\sum \limits_{k=1}^{\infty} (-1)^{k} \frac{x^{2k+1-2}}{(2k+1-2)!}$$
If our index would not start at $k=1$, then our first term would be $x^{2\cdot 0 + 1 - 2} = x^{-1}$. Whoops!
So that is why there is re-indexing. However, we can re-index it back to zero, shifting the index in the exponent and in the factorial in the other direction.
So we almost end up with our initial series
$$\frac{\mathrm{d}^2}{\mathrm{d}x^2}\sin(x)=-\sin(x)=\sum \limits_{k=0}^{\infty} (-1)^{k+1} \frac{x^{2k+1}}{(2k+1)!}$$
Best Answer
All this is saying is that the only difference between $\sum_{k=1}^{\infty}a_k=a_1+a_2+\cdots$ and $\sum_{k=0}^{\infty}=a_0+a_1+a_2+\cdots$ is that the latter has one extra term $a_0$. So $\sum_{k=1}^{\infty}a_k=\left(\sum_{k=0}^{\infty}a_k\right)-a_0$, whatever your sequence $a_k$ is.
(Writing $a_0$ as a sum from $0$ to $0$ might seem obfuscated, but the point is that if you were instead comparing $\sum_{k=10}^{\infty}a_k$ and $\sum_{k=0}^{\infty}a_k$ there would be several extra terms, and it is easier to write the extra as $\sum_{k=0}^9a_k$.)
Here the extra term is $\frac{(-1)^0(\pi/3)^0}{0!}=1$.