Is it possible to derive the nth derivative of$~\exp\left(x\right)\sin^{}\left(x\right)~$using binomial coefficient$~{n\choose k}~$?

Question

I assume$~n\in\mathbb{N}_{\geq0}~$is held.

$$y=\exp\left(x\right)\sin^{}\left(x\right)$$

$$\left(f\cdot g\right)^{\left(n\right)}=\sum_{k=0}^{n}{n\choose k}g^{\left(k\right)}f^{\left(n-k\right)}$$

\begin{equation*}%uasge:&smth\\. . Dont write symbol of line break at the end of row \begin{cases} f:=\sin^{}\left(x\right)\\ g:=\exp\left(x\right)\\ \end{cases} \end{equation*}

$$f^{\left(i\right)}=\sin^{}\left(x+\frac{i\pi}{2}\right)$$

$$g^{\left(i\right)}=\exp\left(x\right)$$

$$\frac{d^{n}y}{dx^{n}}=\left(f\cdot g\right)^{\left(n\right)}=\sum_{k=0}^{n}{n\choose k}\exp\left(x\right)\sin^{}\left(x+\frac{k\pi}{2}\right)$$

$$=\exp\left(x\right)\underbrace{\sum_{k=0}^{n}\sin^{}\left(x+\frac{k\pi}{2}\right){n\choose k}}_\text{How can I handle this?}$$

$$=\exp\left(x\right)\underbrace{\left(\sqrt{2}\right)^{n}\sin^{}\left(x+\frac{n\pi}{4}\right)}_\text{I want to derive this}$$

By the way, I know the following equation. It may not to be applicable here though.

$$\sum_{k=0}^{n}{n\choose k}=2^{n}$$

ADD

I've been looking this post

$$\underbrace{\left(i+1\right)^{n}=\sum_{k=0}^{n}{n\choose k}i^{n}}_\text{This has been on that post}~~\leftarrow~~i~\text{is an imaginary number}$$

The binomial theorem is as below one.

$$\left(a+b\right)^{n}=\sum_{k=0}^{n}{n\choose k}a^{k}b^{n-k}$$

So in this case,$~a=i,b=1~$can be thought.

$$\therefore~~\left(i+1\right)^{n}=\sum_{k=0}^{n}{n\choose k}i^{n-k}\cdot 1^{k}=\sum_{k=0}^{n}{n\choose k}i^{k}\cdot 1^{n-k}$$

$$=\sum_{k=0}^{n}{n\choose k}i^{n-k}=\sum_{k=0}^{n}{n\choose k}i^{k}$$

So I thought the index on the right shoulder should be$~k~$,not$~n~$

Or the following equation can be held?

$$\sum_{k=0}^{n}{n\choose k}i^{k}=\sum_{k=0}^{n}{n\choose k}i^{n}$$

By the way, I got the following.

$$\left(i+1\right)^{n}=\left(\sqrt{2}\right)^{n}\left(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}}\right)^{n}$$

$$=\left(\sqrt{2}\right)^{n}\left(\cos^{}\left(\frac{\pi}{4}\right)+i\sin^{}\left(\frac{\pi}{4}\right)\right)^{n}$$

$$=\left(\sqrt{2}\right)^{n}\left(\cos^{}\left(\frac{n\pi}{4}\right)+i\sin^{}\left(\frac{n\pi}{4}\right)\right)$$

$$=\left(\sqrt{2}\right)^{n}\cdot\exp\left(\frac{i n\pi}{4}\right)$$

Answer 1

$$ \big(\exp(x)\sin x\big)^{(n)}=\sum_{k=0}^n\binom{n}{k}\exp^{(n-k)}(x)\sin^{(k)}(x)=\exp(x)\sum_{k=0}^n\binom{n}{k}\sin(x+k\pi/2) \\=\exp(x)\sum_{k=0}^n\binom{n}{k}\mathrm{Im}\exp(ix+ik\pi/2) =\exp(x)\,\mathrm{Im}\exp(ix)\sum_{k=0}^n\binom{n}{k}\exp(ik\pi/2) \\=\exp(x)\,\mathrm{Im}\exp(ix)\sum_{k=0}^n\binom{n}{k}\exp^k(i\pi/2)=\exp(x)\,\mathrm{Im}\exp(ix)\sum_{k=0}^n\binom{n}{k}i^k\\=\exp(x)\,\mathrm{Im}\exp(ix)(1+i)^n $$

Another possibility which avoids Leibniz's Rule $$ \exp(x)\sin x=\mathrm{Im}\,\big(\exp(x) \exp(ix)\big) =\mathrm{Im}\,\big(\exp(x+ix)\big) $$ So $$ \big(\exp(x)\sin x\big)^{(n)} =\mathrm{Im}\,(1+i)^n\big(\exp(x+ix)\big)= $$ But $1+i=2^{1/n}\exp(i\pi/4)$ and hence $$ \big(\exp(x)\sin x\big)^{(n)} =\mathrm{Im}\,(1+i)^n\big(\exp(x+ix)\big)=2^{n/2}\mathrm{Im}\,\big(\exp(n\pi/4)\exp(x+ix)\big)\\=2^{n/2}\mathrm{Im}\,\big(\exp(in\pi/4)\exp(x+ix)\big)=2^{n/4}\exp(x)\sin(x+n\pi/4) $$