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

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

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 $$ \big(\exp(x)\sin x\big)^{(n)} =\mathrm{Im}\,(1+i)^n\big(\exp(x+ix)\big)=2^{n/2}\mathrm{Im}\,\big(\exp(n\pi/4)\exp(x+ix)\big)\\=2^{n/2}\mathrm{Im}\,\big(\exp(in\pi/4)\exp(x+ix)\big)=2^{n/4}\exp(x)\sin(x+n\pi/4) $$