In Axler's Linear Algebra Done Right we are asked to verify the assertion that the list $1,z,...,z^m$ is linearly independent (l.i) in $P(\pmb{\mathbb{F}})$ for each nonnegative integer $m$.
The way I showed this was to assume these vectors of $P(\pmb{\mathbb{F}})$ are linearly dependent (l.d.). That is, there is some linear combination of these vectors, with coefficients not all zero, that is the zero polynomial.
Let such a polynomial be
$$p(z)=\lambda_0+\lambda_1z+...+\lambda_mz^m=0$$
which is true for $\lambda_i$ not all zero and for all $z$.
Then, $z=0$ implies $p(0)=\lambda_0=0$.
Since polynomials are differentiable everywhere, we can compute the derivative at $z=0$
$$p'(0)=\lambda_1=0$$
and we can keep doing this to find that all the $\lambda$'s are zero.
This result contradicts our assumption that these coefficients were not all zero.
Hence, we can infer that in fact our initial vectors $1,z,...,z^m$ are l.i.
Questions
Is this proof correct?
What are some alternative ways of proving the result in question?
So what you have written as $P (F)$ is commonly written $F[z]$ the free $F$ algebra with $z$ for generator. It is more commonly called the ring of univariate polynomials. You proof idea is good since you want to use induction, but the derivation is not sufficient; usual fields like $\mathbb{Q}, \mathbb{R}$ or $\mathbb{C}$ have zero as a characteristic, which means the morphism $\mathbb{Z} \to F$ that sends $1$ to $1_F$ is injective. For some fields like finite fields this is not the case, we say that the characteristic is positive.
An example would be $\mathbb{F}_2$, the field of integers modulo $2$, if you multiply anything by $2$ it's the same as multiplying by $0$ since $2 = 0$ in this field. Then if you derivate $z^2$, you get $2z = 0$, but you don't have $x^2 = k \in \mathbb{F}_2$...
Now lets correct this proof; our induction property is the following: For $n \in \mathbb{N}$, the family $1, \dots, z^n$ if free.
for $n=0$ it's clear
now if it has been verified for $n-1$, take
$$\lambda_0 + \lambda_1 z + \cdots + \lambda_n z^n$$ By evaluating in $0$ you get $\lambda_0 = 0$ this gives you $$z(\lambda_1 + \cdots + \lambda_n z^{n-1} ) = 0$$ Since $F[z]$ s a domain and $z \neq 0$, we have $\lambda_1 + \cdots + \lambda_n z^{n-1} = 0$. By induction, $\lambda_1 = \cdots = \lambda_n = 0$
If you already had this result, you could have stated by uniqueness of the coefficients of a polynomial, $\lambda_0 = \cdots = \lambda_n = 0$. But the method here works for any family graded in degree.