Let $K$ be a field of characteristic $p \geq 0$ and let $M$ be a matrix $n \times n$ over $K$. If $p \nmid n$ and $Tr(M^i) = 0$ for all $i = 1,\dots,n$, how to prove that $M + Id_n$ is invertible?
If $p=0$, we can write down a proof using the usual general formula to characteristic polynomial of a matrix (that is an application of Newton's identities).
You cannot, because your assertion is false. Consider $n=3,\ p=2$ and $M=\operatorname{diag}(1,1,0)$. For any $i\ge1$, we have $M^i=M$ and $\operatorname{tr}(M^i)=\operatorname{tr}(M)=0$. Yet $M+I=\operatorname{diag}(0,0,1)$ is not invertible.
Your assertion is true, however, if you replace the condition $p \nmid n$ by $p=0$ or $p>n$. To prove the assertion, you may modify JBC's argument to a related question. The key is that $M$ must be nilpotent. Suppose the contrary. Then $M$ has, over the algebraic closure of $K$, a set of distinct non-zero eigenvalues $\lambda_1,\ldots,\lambda_r$. Let $n_i$ the multiplicity of $\lambda_i$. From the condition that $\operatorname{tr}(M^i)=0$ for $i=1,2,\ldots,n$, we get $$ \underbrace{\pmatrix{ \lambda_1&\lambda_2&\cdots&\lambda_r\\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2\\ \vdots & \vdots & \vdots & \vdots\\ \lambda_1^r & \lambda_2^r & \cdots & \lambda_r^r }}_V \pmatrix{n_1 \\ n_2 \\ \vdots \\ n_r} =\pmatrix{0 \\ 0\\ \vdots \\ 0}.\tag{1} $$ The square matrix $V$ on the left is equal to $$ \pmatrix{ 1 & 1 & \cdots & 1 \\ \lambda_1&\lambda_2&\cdots&\lambda_r\\ \lambda_1^2 & \lambda_2^2 & \cdots & \lambda_r^2 \\ \vdots & \vdots & \vdots & \vdots \\ \lambda_1^{r-1} & \lambda_2^{r-1} & \cdots & \lambda_r^{r-1} } \pmatrix{\lambda_1\\ &\lambda_2\\ &&\ddots\\ &&&\lambda_r}, $$ which is the product of a Vandermonde matrix and a diagonal matrix. As all $\lambda_i$s are nonzero and distinct, $V$ is invertible. Yet, as $p>n$, the vector $(n_1,\ldots,n_r)^T$ is nonzero. So, we arrive at a contradiction in $(1)$. Hence $M$ must be nilpotent and in turn $M+I$ is invertible.
For the matrix $M$ show that all eigenvalues are 0, that is nilpotent. Then it follows that $M+\mathrm{Id}_n$ has all eigenvalues 1, and hence invertible.
