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 $c:=1+\frac2{\sqrt3}.$

To illustrate the accuracy of the upper bound $B_n$ on $|I_{n,k}|$, here is the graphs $\{(n,k,r_{n,k})\colon n=2,\dots,30,\ k=1,\dots,n\}$ and $\{(n,r_{n,n-1})\colon n=2,\dots,30\}$ for $r_{n,k}:=\dfrac{|I_{n,k}|}{B_n}$, which suggest that that $r_{n,k}$ is not less that something like $0.3$ for natural $n>k$.

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 $c:=1+\frac2{\sqrt3}.$

To illustrate the accuracy of the upper bound $B_n$ on $|I_{n,k}|$, here are the graphs $\{(n,k,r_{n,k})\colon n=2,\dots,30,\ k=1,\dots,n\}$ and $\{(n,r_{n,n-1})\colon n=2,\dots,30\}$ for $r_{n,k}:=\dfrac{|I_{n,k}|}{B_n}$, which suggest that $r_{n,k}$ is not less than something like $0.3$ for natural $n>k$.

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 $c:=1+\frac2{\sqrt3}.$

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 $c:=1+\frac2{\sqrt3}.$

To illustrate the accuracy of the upper bound $B_n$ on $|I_{n,k}|$, here is the graphs $\{(n,k,r_{n,k})\colon n=2,\dots,30,\ k=1,\dots,n\}$ and $\{(n,r_{n,n-1})\colon n=2,\dots,30\}$ for $r_{n,k}:=\dfrac{|I_{n,k}|}{B_n}$, which suggest that that $r_{n,k}$ is not less that something like $0.3$ for natural $n>k$.