2) => 1)
Let $L/K$ be an algebraic extension and $\alpha \in L$. Let $f$ be the minimal polynomial of $\alpha$ over $K$. We now that there is some minimal $n$ such that $f(x^{p^{n}})$ is separable and irreducible. We want to show that $n=0$. Let $f(x)=a_n x^n + \dots a_0$, then choose $b_i \in K$ such that $(b_i)^{p^n}=a_i$ (using the surjectivity of the Frobenius), then we get that if we consider $g(x)=b_nx^n+ \dots b_0$, $g^{p^n}=f(x^{p^n})$ which contradicts that $f(x^{p^n})$ is irreducible unless $n=0$

@MatheinBoulomenos I thanked you for this answer: " $\Lambda(V \oplus W) \cong \Lambda(V) \otimes \Lambda(W)$ (taking the graded tensor product), this holds for vector spaces, but it also carries over to bundles. Now if we take the projections $p_1: M \times N \to M$ and $p_2:M \times N \to N$, then we have $T(M \times N) \cong p_1^*TM \oplus p_2^* TN$, so we get $\Lambda(T(M \times N)) \cong \Lambda(p_1^* TM) \otimes \Lambda(p_2^* TN)$, so looking at the componenet in degree $m+n$,
we get $\Lambda^{n+m}(T(M \times N)) \cong \Lambda^{m}(p_1^*TM) \otimes_{\Bbb R} \Lambda^{n}(p_2^*TN)$"