Lemma 6.1.2 in Herstein's 'Topics in Algebra' 'Let $A$ be an algebra, with unit element, over $F$, and suppose that $A$ is of dimension $m$ over $F$. Then every element in $A$ satisfies some non-trivial polynomial in $F[x]$ of degree at most $m$.' (pp. 263)
The proof goes on to claim that any selection of $m + 1$ elements, say, $e, a, a^2, ..., a^m$, in $A$, will be linearly dependent. That much is fine. Because of this, they must satisfy some polynomial $\alpha_0 + \alpha_1x + ... + \alpha_mx^m = 0$, where at least one of the coefficients is non-zero. 
This is all good. My misunderstanding is this: why does the lemma restrict the degree of the polynomial to at most $m$? Can I not apply the same argument to $m+2$ elements or more? Why not?
 A: You are right , in that the argument could have been applied to the elements $e,a,a^2,...,a^{k}$ for any $k \geq m$ and we would have a linear combination of these elements summing to zero, which would give a polynomial of degree $k$ satisfied by $a$. 
However, note that the dimension of $A$ over $F$ is $m$, so while it is useful, that every element of $a$ satisfies some polynomial over $F$, the statement that such a polynomial has to have degree at most $m$ is a very strong one, because then we even have a hold on the degree of the polynomial satisfied by $a$ : initially it could have been any natural number, now it is something less than $m$. Remember, polynomials of small degree are (generally) easier to work with than polynomials of large degree. For example, if $m=2$ then it would be nice to know that every element is the root of some quadratic polynomial, rather than say a polynomial of degree $74$ or something.
To show this strengthened fact, we must restrict ourselves to $k=m$ above. As a remark, simply by multiplying by $x$, we can make a polynomial of degree $m$ satisfied by $a$, into a polynomial of degree $m+1$ satisfied by $a$. So the point is that we are anyway in trivial territory with the argument above applied to $k > m$.
