For the first statement, if $n=2$, let $A=\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2}\end{pmatrix}$ and $B= \begin{pmatrix} 1 & 1 \\ -1 & -1\end{pmatrix}$
Then
$AB-BA= \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2}& \frac{1}{2}\end{pmatrix} - \begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2}& +\frac{1}{2}\end{pmatrix}= \begin{pmatrix} 0 & 1 \\ 1& 0\end{pmatrix}$
I think this can be generalized to $2m \times 2m$ matrices by using
$\begin{pmatrix} A & 0 & 0& ..&0 \\ 0 & A & 0& ..&0 \\ . & .& . & ..& . \\ 0 & 0 & 0& ..&A \\\end{pmatrix}$ and $\begin{pmatrix} B & 0 & 0& ..&0 \\ 0 & B & 0& ..&0 \\ . & .& . & ..& . \\ 0 & 0 & 0& ..&B \\\end{pmatrix}$
For the second statement $A=I$...
If the problem asks instead to prove that
$tr(A)^n \geq n^n \det(A)$
then let $\lambda_1,..,\lambda_n$ be the eigenvalues. Then they are real and positive (WHY?), thus by $AM-GM$ we have:
$\frac{tr(A)}{n}=\frac{\lambda_1+...+\lambda_n}{n} \geq \sqrt[n]{\lambda_1 \cdot ... \lambda_n}=\sqrt[n]{\det(A)}$