Let $N \in M_n(\mathbb{F})$ be nilpotent.
Prove that for any $1 \leq k \in \mathbb{N}$ a matrix $B \in M_n(\mathbb{F})$ exists such that $B^k=I+N$
I have no idea how to het started here. We've just covered Rational form and JCF. Any ideas? Suggestions?
Assuming that you are working over a field of characteristic$~0$, you can use the binomial formula for exponent $\frac1k$. That is, as a series in $N$, which terminates because $N$ is nilpotent, one has that $$ B=(I+N)^{\frac1k}\buildrel{\rm def}\over=\sum_{i\in\Bbb N}\binom{1/k}iN^i $$ is a solution to $B^k=I+N$. Moreover this is the unique solution that is a polynomial in $N$.
Here one takes by definition $\binom xi=\frac{x(x_1)\ldots,(x-i+1)}{i!}$, which is used above for $x=\frac1k$. Given the size of the nilpotent matrix$~N$, the sum over $i$ is equivalent to the finite sum $\sum_{i=0}^{n-1}$.
Without the hypothesis of characteristic$~0$ a solution does not always exist. For a simple case, take $k=p=\operatorname{char}(\Bbb F)$ and write $B=I+A$, then $B^k=(I+A)^p=I+A^p$, so one has to solve $A^p=N$, but this has no solution if $N$ has a single Jordan block of size $n>1$. (It will have a solution only for very few nilpotent matrices: $A$ must be nilpotent, and the size of each Jordan blocks of $A$ is split into $p$ almost equal-size pieces giving sizes of Jordan blocks of $N$; this is a severe constraint on the Jordan type of $N$.)