If $(T-\lambda)^k(v) =0 $ and $(S-\mu)^l(w)= 0$, then $(T\otimes S - \lambda\mu)^{k+l}(v\otimes w) = 0$. 
Let $T\in \mathcal L(V,V)$ and $S\in \mathcal L(W,W).$ If $k,l\in \mathbb Z_{>0}, \lambda, \mu\in \mathbb F, v\in V, w\in W$ satisfies:  $(T-\lambda I_V)^k(v) =0 $ and $(S-\mu I_W)^l(w)= 0$, then  $(T\otimes S - \lambda\mu I_V\otimes I_W)^{k+l}(v\otimes w) = 0$.

I'm having a hard time trying to prove this statement. I'll post here my progress:
We evaluate our function: $(T\otimes S - \lambda\mu)$ at $(v\otimes w)$:
$$(T\otimes S-\lambda\mu)(v\otimes w) = T\otimes S(v\otimes w) -\lambda\mu(v\otimes w) \\ = T(v)\otimes S(w) - \lambda\mu(v\otimes w).$$
Evaluating again, we have:
$$(T\otimes S-\lambda\mu)(T(v)\otimes S(w) - \lambda\mu(v\otimes w)) = \\= T\otimes S(T(v)\otimes S(w)) - T\otimes S(\lambda\mu(v\otimes w)) - (\lambda\mu) T(v)\otimes S(w) - (\lambda\mu)^2(v\otimes w)=\\= T^2(v)\otimes S^2(w) - \lambda\mu(T(v)\otimes S(w))-\lambda\mu(T(v)\otimes S(w) - (\lambda\mu)^2(v\otimes w).$$
I can't proceed from those results, that is, it's not clear what will be $(T\otimes S - \lambda\mu)^{k+l}(v\otimes w)$. 
Any help wou be much appreciated. 
 A: One next step that you could take in your final line: combine like terms. So you have 
$$T^2(v)\otimes S^2(w) - \lambda\mu(T(v)\otimes S(w))-\lambda\mu(T(v)\otimes S(w)) - (\lambda\mu)^2(v\otimes w)\\= T^2(v)\otimes S^2(w) - 2\lambda\mu(T(v)\otimes S(w)) - (\lambda\mu)^2(v\otimes w).$$
Perhaps this suggests where this can go in general. Because $T\otimes S$ commutes with the scalar multiplication by $\lambda\mu$, we may use the binomial theorem
$$
(T\otimes S - \lambda\mu)^{k+l}(v\otimes w)=\sum^{k+l}{k+l\choose j}\lambda^{k+l-j}\mu^{k+l-j} T^j(v)\otimes S^j(w).
$$
The identity $(x+y)^k(x+y)^l=(x+y)^{k+l}$ yields an identity on binomial coefficients which we can then use to finish the problem.
In detail, the identity we seek is called Vandermonde's identity. We have
$$
(x+y)^k(x+y)^\ell=\left(\sum^k_i{k\choose i}x^iy^{k-i}\right)\left(\sum^\ell_j{\ell\choose j}x^jy^{\ell-j}\right)\\=\sum^k_i\sum^\ell_j{k\choose i}{\ell\choose j}x^{i+j}y^{k+\ell-i-j}.
$$
Now a double sum is the sum of the elements of a rectangular $(i,j)$-array. It is often helpful to reindex to view it as a sum of diagonals of fixed $i+j$. That's clearly what we want here, as we need to combine like terms to compare our sum with the binomial expansion for $(x+y)^{k+\ell}.$ So let's put $n=i+j$ and put $N=k+\ell$ and rewrite our doublesum.
$$
\sum^k_i\sum^\ell_j{k\choose i}{\ell\choose j}x^{i+j}y^{k+\ell-i-j} =\sum_{i+j}^{k+\ell}\sum_{i+j\text{ fixed}}{k\choose i}{\ell\choose j}x^{i+j}y^{k+\ell-i-j}\\
=\sum_{i+j}^{k+\ell}\left(\sum_{i+j\text{ fixed}}{k\choose i}{\ell\choose j}\right)x^{i+j}y^{k+\ell-i-j} = \sum_{n}^{N}\left(\sum_{i}^n{k\choose i}{N-k\choose n-i}\right)x^{n}y^{N-n}.
$$
But the binomial theorem also tells us that
$$
(x+y)^N=\sum_n^N{N\choose n}x^ny^{N-n}.
$$
Since $(x+y)^k(x+y)^\ell=(x+y)^{k+\ell}=(x+y)^N,$ we may equate these two sums, polynomials in $x$ and $y$ which are equal if and only if the coefficients are equal termwise (and this is the reason we combined like terms in the first sum). So we conclude that
$$
\sum_{i}^n{k\choose i}{N-k\choose n-i} = {N\choose n}.
$$
This concludes the derivation of Vandermonde's identity. You can also find a proof on Wikipedia, as well as here on m.se here or here. Or you may simply take the identity on faith, and use it.
So let's get back to the problem at hand. We basically reproduce the derivation of the Vandermonde identity in reverse. We have
$$
(T\otimes S - \lambda\mu)^{k+\ell}(v\otimes w)=\sum^{k+\ell}_n{k+\ell\choose n}\lambda^{k+\ell-n}\mu^{k+\ell-n} T^n(v)\otimes S^n(w)\\
=\sum^{k+\ell}_n\left(\sum {k\choose n}{\ell\choose k-n}\right)\lambda^{k+\ell-n}\mu^{k+\ell-n} T^n(v)\otimes S^n(w)\\
=(T-\lambda)^k\otimes (S-\mu)^\ell(v\otimes w).
$$
And the latter vanishes by hypothesis.
