Showing that the set of integers that can be expressed as tenth powers of integers is countable I'm trying to show that the set of all integers which are tenth powers of integers. Let $X = \{m \in \mathbb{Z} : \text{$m=k^{10}$ for some k}  \in \mathbb{Z}\}$. We want to show that $X$ is countable, i.e. that there exists an injection from $X$ to $\mathbb{N}$. I know that $X -\{0\} \subset \mathbb{N}$ so I extended the inclusion function from $X-\{0\}$ to $\mathbb N$ in order to construct an injective map. This is the function I came up with: $$f: X \rightarrow \mathbb N, x \mapsto f(x) = \begin{cases}
 x&\text{if}\, x\ne 0\\
 2&\text{if}\, x=0\\
\end{cases}$$
Now proving that this is an injection is quite trivial but I was wondering if there are any more easier injections from $X$ to $\mathbb N$.
 A: "Now proving that this is an injection is quite trivial but I was wondering if there are any more easier injections from X to N."
Well $x\mapsto x + 1$.
Well, my first thought would be $f(x) = \sqrt[10]{x} + 1$. Not only an injection but a bijection.
IMO you are giving much more thought to this than it deserves.  Simply noting $X \subset \mathbb Z$ and $\mathbb Z$ is countable is enough.  So  as $\mathbb Z$ is countable, an injection into $\mathbb Z$ is as acceptable as an injection into $\mathbb N$.
After all, if $\phi: \mathbb Z\to \mathbb N$ is a bijection then the $\phi: X \subset \mathbb Z \to \phi(X)\subset \mathbb N$ is an injection.
(Which you could use to say $\phi: x\mapsto 2|x|$ if $x > 0$ and $\mapsto 2|x|+1$ if $x < 0$, so an natural injection would be $x \to 2x$ if $x > 0$ and $x\to 1$ if $x = 0$.  Which is ... silly.) 
A: A simple injection $X \to \mathbb N$ is given by $x \mapsto 2^x$ (a kind of Gödel numbering).
A: Mapping each element of $X$ to its unique positive tenth root is an injection. But @lulu 's comment is a better reason: any infinite subset of a countable set is countable, hence the existence of a bijection.
