Recall that the free associative $k$-algebra on a set $X$ has a unique $\mathbb N_0$-grading as a $k$-algebra $k\langle X\rangle = \bigoplus_{n \geq 0} k\langle X\rangle^{(n)}$, where the image of $X$ in $k\langle X \rangle$ is of degree 1.
I am confused what "degree" means in this context, and more importantly what it tells us. I don't have any kind of definition written in my notes unfortunately. Any help would be very appreciated, thanks in advance.
Recall that a graded $k$-algebra, $R$, is a $k$-algebra that decomposes as an abelian group as the direct sum $$R=\bigoplus_{i=0}^\infty R_i$$ for some subgroups $R_i$ such that $R_iR_j\subseteq R_{i+j}$ for all $i$ and $j$ (usually we want that $R_0=k$).
The homogeneous elements of $R$ are the elements of the groups $R_i$, and the degree of a nonzero homogeneous element $x\in R_i$ is $i$, the index of the subgroup that it is an element of.
In this particular case, the groups $R_n$ are generated by the words of length $n$, so the degree of a word $x_1x_2\cdots x_n$ is $n$. Since $X$ is not strictly speaking a subset of $k\langle X\rangle$, they are saying that the image of any element $x\in X$ in $k\langle X\rangle$ has degree 1, i.e., it maps to the word of length one, $x$.