= Solution
Work with a nonsingular <multivariate normal distribution>, so $V$ is <positive-definite>. The claimed finite <maximum-likelihood estimates> also require the centered <sample covariance matrix> to be nonsingular; the exceptional case is discussed below.
Independence gives, up to a constant, the <log-likelihood>
$$
\ell(\mu,V)=-\frac n2\log\det V
-\frac12\sum_{i=1}^n(Y_i-\mu)^TV^{-1}(Y_i-\mu).
$$
Let $\bar Y=n^{-1}\sum_iY_i$ and $S=n^{-1}\sum_i(Y_i-\bar Y)(Y_i-\bar Y)^T$. The centered sum is zero, and expansion therefore gives
$$
\sum_i(Y_i-\mu)(Y_i-\mu)^T
=nS+n(\bar Y-\mu)(\bar Y-\mu)^T.
$$
For every fixed <positive-definite matrix> $V$, the extra quadratic term in the <log-likelihood> is $-n(\bar Y-\mu)^TV^{-1}(\bar Y-\mu)/2$, with unique maximum at $\mu=\bar Y$.
Assume $S$ is <positive-definite>. Maximizing over $V$ now means minimizing $\log\det V+\operatorname{tr}(V^{-1}S)$. Put $A=S^{1/2}V^{-1}S^{1/2}$, another <positive-definite matrix>. Then
$$
\log\det V+\operatorname{tr}(V^{-1}S)
=\log\det S-\log\det A+\operatorname{tr}A.
$$
If $a_1,\ldots,a_p>0$ are its <eigenvalues>, the variable part is $\sum_j(a_j-\log a_j)$. The scalar inequality $a-\log a\ge1$, with equality only at $a=1$, shows that this is minimized uniquely at $A=I$, or $V=S$. Thus
$$
\boxed{\widehat\mu=\bar Y,\qquad
\widehat V=S=\frac1n\sum_i(Y_i-\bar Y)(Y_i-\bar Y)^T.}
$$
The divisor is $n$ because this is <maximum likelihood estimation>, not the unbiased <covariance> estimate with divisor $n-1$.
The nonsingularity condition matters. The centered scatter has <rank> at most $n-1$. If $S$ is singular, choose $\mu=\bar Y$, retain positive <covariance> <eigenvalues> on its range and let a <covariance> <eigenvalue> on its <null space> tend to zero. The quadratic <likelihood> term remains finite, while $-\log\det V\to+\infty$. This is <singular covariance and nonexistence of a Gaussian maximum likelihood estimate>: there is no finite maximizer over nonsingular $V$. Under a nonsingular <Gaussian> population, $S$ is positive definite almost surely when $n>p$; for $n\le p$ it is necessarily singular. The asserted estimate and the inverse in the next part have their usual meaning under $n>p$.
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