I want to prove the following inequality: $$E\left[\left|\int_{t0}^t f(s,\omega)dW_s \right|^{2n}\right]\le (t-t_0)^{n-1}[n(2n-1)]^n\int_{t0}^tE[|f(s,\omega)|^{2n}]ds$$
I have been trying with Ito-Isometry and Cauchy Schwarz Inequality to get the Expectation under the integral. Where does the factor $[n(2n-1)]^n$ come from? Applying Ito Formula:
$$E[\lvert\int_{t_0}^t f(s,\omega)dW_s\rvert^{2n}]=E[\int f^2(t,\omega)*n(2n-1)|\int f(s,\omega)dW_s|^{2n-2}dt]$$ Repeating the recursion n times we will get $$E[\lvert\int_{t0}^t f(s,\omega)dW_s \rvert^{2n}]= E[\int f^{2n}(t,\omega)*n(2n-1)*(n-1)*(2n-3)*.....*2 dt]\le E[\int f^{2n}(t,\omega)*n(2n-1)^n dt]\le n(2n-1)^nE[\int |f|^{2n}(t,\omega) dt]$$
Now how does the $(t-t_0)^{n-1}$ comes in?
If we set $$X_t := \int_{t_0}^t f(s) \, dW_s$$ then Itô's formula shows
$$\mathbb{E} \left( \left| \int_{t_0}^t f(s) \, dW_s \right|^{2n} \right) = \mathbb{E}(|X_t|^{2n}) = n (2n-1) \mathbb{E} \left( \int_{t_0}^t |X_s|^{2n-2} f(s)^2 \, ds. \right) \tag{1}$$
Because of the additional factor $f(s)^2$ on the right-hand side we cannot simply iterate this argumentation (because then the stochastic integral doesn't vanish). Instead, we have to apply Hölder's inequality:
$$\mathbb{E}(f(s)^2 |X_s|^{2n-2}) \leq \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \bigg[ \mathbb{E}(|X_s|^{2n}) \bigg]^{1-1/n}.$$
Since $|X_s|^{2n}$ is a submartingale, we have $\mathbb{E}(|X_s|^{2n}) \leq \mathbb{E}(|X_t|^{2n})$ for all $s \in [t_0,t]$. Combining both inequalities with $(1)$, we get
$$\mathbb{E}(|X_t|^{2n}) \leq n(2n-1) \bigg[\mathbb{E}(|X_t|^{2n}) \bigg]^{1-1/n} \int_{t_0}^t \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \, ds.$$
Hence,
$$\mathbb{E}(|X_t|^{2n}) \leq (n(2n-1))^n \left( \int_{t_0}^t \bigg[ \mathbb{E}(|f(s)|^{2n}) \bigg]^{1/n} \, ds \right)^n.$$
Applying Jensen's inequality proves the assertion.
Remark: There is a similar inequality for the running maximum: $$\mathbb{E} \left( \sup_{t_0 \leq s \leq t} |X_s|^{2n} \right) \leq \left( \frac{2n}{2n-1} \right)^{2n} (t-t_0)^{n-1} (n(2n-1))^n \int_{t_0}^t \mathbb{E}(|f(s)|^{2n}) \, ds;$$ this is a direct consequence of the Burkholder-Davis-Gundy inequality.