We want to upper-bound $|I_{n,k}|$, where 
$$I_{n,k}:=\int_0^{\pi/3} e^{-itn} (1-e^{it})^k\,dt.$$
Integrating by parts, we have 
$$I_{n,k}=\frac{e^{-itn}}{-in}(1-e^{it})^k\Big|_0^{\pi/3}
-\int_0^{\pi/3} \frac{e^{-itn}}{-in}\,k(1-e^{it})^{k-1}(-ie^{it})\,dt.$$
So, 
$$n|I_{n,k}|\le|1-e^{i\pi/3}|^k+
\int_0^{\pi/3}k|1-e^{it}|^{k-1}\,dt.$$
Note that for $t\in(0,\pi/3)$ we have $|1-e^{it}|=2\sin\frac t2$, and hence $|1-e^{i\pi/3}|=1$ and 
$$\int_0^{\pi/3}k|1-e^{it}|^{k-1}\,dt
=2^k\int_0^{1/2}k s^{k-1}\,\frac{ds}{\sqrt{1-s^2}} \\ 
\le2^k\int_0^{1/2}k s^{k-1}\,\frac{ds}{\sqrt{1-(1/2)^2}}
=\frac1{\sqrt{1-(1/2)^2}}=\frac2{\sqrt3}.$$ 
So, 
$$|I_{n,k}|\le B_n:=\frac cn,$$
where $c:=1+\frac2{\sqrt3}.$