Completeness can be justified indirectly if you invoke results of the Exponential family.

The pdf of $X$ for $\theta\in\mathbb R^p$ is

\begin{align} f_{\theta}(x)&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}\sum_{i=1}^p (x_i-\theta_i)^2\right] \\&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}x^Tx-\frac{1}{2}\theta^T\theta+x^T\theta\right]\quad,\small x=(x_1,\ldots,x_p)\in\mathbb R^p \end{align}

This density is a member of the exponential family, which guarantees that a complete sufficient statistic for $\theta$ is $X^T$, or simply $X$.

Since $X$ is unbiased for $\theta$, it has to be the UMVUE.

Completeness can be justified indirectly if you invoke results of the Exponential family.

The pdf of $X$ for $\theta\in\mathbb R^p$ is

\begin{align} f_{\theta}(x)&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}\sum_{i=1}^p (x_i-\theta_i)^2\right] \\&=\frac{1}{(\sqrt{2\pi})^p}\exp\left[-\frac{1}{2}x^Tx-\frac{1}{2}\theta^T\theta+x^T\theta\right]\quad,\small x=(x_1,\ldots,x_p)\in\mathbb R^p \end{align}

This density is a member of a full rank exponential family, which guarantees that a complete sufficient statistic for $\theta$ is $X^T$, or simply $X$.

I think completeness can also be proved in the following way. Let $g(\cdot)$ be any function of $x$.

Then, $$E_{\theta}[g(X)]=0\quad\forall\,\theta\implies \int_{\mathbb R^p}e^{x^T\theta}g(x)e^{-\frac12x^Tx}\,dx=0\quad\forall\,\theta$$

The above is a (multidimensional) bilateral Laplace transform of $g(x)e^{-\frac12x^Tx}$, which implies $$g(x)e^{-\frac12x^Tx}=0\quad,\text{ a.e.}$$

That is, $$g(x)=0\quad,\text{ a.e.}$$