Let $K/F$ be a field extension, $t,w \in K$ such that $t$ is transcendental over $F$ and $w$ is transcendental over $F(t)$. Prove that for any positive integer numbers $n$ and $m$ the equality $[F(t,w):F(t^n,w^m)]=nm$.
This is what I did:
$$[F(t,w):F(t^n,w^m)]=[F(t,w):F(t,t^n,w^m)]·[F(t,t^n,w^m):F(t^n,w^m)]=[F(t,w):F(t,w^m)]·[F(t,w^m):F(t^n,w^m)]$$
For proving that $[F(t,w):F(t,w^m)]=m$, I use the polynomial
$$p(X)=X^m-w^m\in F(t,w^m)[X]$$ and I want to say it's irreducible over $F(t,w^m)$, $irr(w,F(t,w^m))=X^m-w^m$. It's monic and $w$ is a root of $p(X)$ but I don't know how to see it's irreducible over $F(t,w^m)$ and don't know how to use the fact that $w$ is transcendental over $F(t)$ to obtain the number $m$.
Very good start.
Lemma: If $w$ is transcendental over $L$, then $|L(w):L(w^m)|=m$.
Proof: Clearly $x^m-w^m\in L(w^m)$ is a witness for the degree being at most $m$.
We show that there is no nonzero polynomial $f(x)\in L(w^m)$ of degree $r$ less than $m$ such that $w$ is a root of $f$. Given such an $f$, its coefficients (elements of $L(w^m)$) can be expressed as rational functions of $w$ over $L$. We may multiply by all the denomiators, and obtain an $f$ such that all coefficients are polynomials of $w^m$ over $L$. So $f(x)= \sum\limits_{i=0}^{r} g_i(w^m)x^i$. Let $a_sw^{ms}$ be the leading term of $g_r(w^m)$. Substitute $w$, and collect like terms to obtain a polynomial in $L(w)$. But nothing can cancel the monomial $a_sw^{ms+r}$, as the only term of the form $cw^{ms+r}$ comes from $a_sw^{ms}w^r$. Indeed, surely nothing can cancel it in the product $g_r(w^m)w^r$, i.e., when we substitute into the main term. As for the other terms, the polynomial coefficient $g_i(w^m)$ consists of terms where the power of $w$ is divisible by $m$, and the $x$-power has exponent between $0$ and $r-1$. So the powers of $w$ that you obtain after the substitution $x=w$ and multiplication all have exponents incongruent to $r$ $\pmod m$ in those terms (this is the point where $r<m$ is used). Thus we obtain a nonzero polynomial of $w$ over $L$, and hence $f(w)\neq 0$.
Using this lemma, clearly $|F(t)(w):F(t)(w^m)|=m$ by $L=F(t)$. Somewhat less obviously, but we also obtain $|F(w^m)(t):F(w^m)(t^n)|=n$ by $L=F(w^m)$. In order to use the lemma, just show that $t$ is also transcendental over $F(w^m)$. (Use proof by contradiction and the fact that $w$ is transcendental over $F(t)$.)