Covariance between Wiener process and stochastic integral I want to calculate $cov($$w_t$ , $\int_0^{t}s^{n}dw_s)$ .
I have tried integration by parts:
$$\int_0^{t}s^{n}dw_s = t^{n}w_{t} - n\int_{0}^{t}s^{n-1}w_sds$$ Further, I think to use this formula $n$ times, but I can't do it on this step. My idea was to get the equation at the end that will contain next components (maybe with some coefficients): $\int_0^{t}sdw_s$, $\int_0^tw_sds$, $w_t$ and the component that if I count a mathematical expactation from it I will get the

$cov($$w_t$, $\int_0^{t}s^{n}dw_s)$ (the mathemetical expectation for the first three components I know). And from this equation I would count the covariance, but I can't understand how I can continue. I'll be happy for any idea.
The most clever way is, as @Did suggested, to use Itô's isometry. However, you can also use the integration by parts formula:

$$\int_0^t s^n \, dW_s = t^n W_t - n \int_0^t s^{n-1} W_s \, ds.$$

Multiplying both sides by $W_t$ and taking the expectation we get

$$\begin{align*} \mathbb{E} \left( W_t \cdot \int_0^t s^n \, dW_s \right) &amp;= t^n \underbrace{\mathbb{E}(W_t^2)}_{t} - n \int_0^t s^{n-1} \underbrace{\mathbb{E}(W_s W_t)}_{\min\{s,t\}=s}\,d s \\ &amp;= t^{n+1} - n \int_0^t s^n \, ds =\frac{1}{n+1} t^{n+1}.\end{align*}$$
The solution is
$$cov(w_t, \int_0^ts^ndw_s)=E(w_t - Ew_t)(\int_0^ts^ndw_s- E\int_0^ts^ndw_s)=E(w_t\int_0^ts^ndw_s-w_tE\int_0^ts^ndw_s)=\int_0^ts^nds-Ew_tE\int_0^ts^ndw_s=\int_0^ts^nds=\frac{t^{n+1}}{n+1}$$