Let $G_m= k^{*}=k-{0}$ be the multiplicative group. We know this is an Algebraic group also. How does one prove any algebraic group morphism $G_m \rightarrow G_m$ is of the form $t \mapsto t^{n}$ for some $n \in \mathbb Z$.
Note that an algebraic group morphism is a morphism of varieties which is also a group homomorphism.
On
Lets assume that $k$ is infinite. If $f(x)$ is a polynomial such that $f(st)=f(s)f(t)$ all $s,t \in k^*$, then $f(tx)=f(t)f(x)$ holds for all nonzero $x$, but if $k$ is infinite this implies it holds for all $x$. Therefore $f(0)=f(t)f(0)$ so either $f$ is constant (=1) or $f(0)=0$, and $f$ has no constant term. Now one can assume that $f(x)=x^mg(x)$ and repeat the same argument for $g$. This also works for rational functions where if we assume that the numerator and denominator have nonzero constant coefficients a contradiction results.
Let $\varphi:G_m\to G_m$ be defined by $t\mapsto\frac{f(t)}{g(t)}$, where $f,g$ are polynomials. Assume $k$ is algebraically closed. If one of $f,g$ is not $t^n$ for some $n$,then it maps a nonzero element in $k$ to $0$. Thus for $\varphi$ to be well defined, the nonzero roots of $f$ and $g$ must be the same, meaning that $\varphi(t)=t^l$ for some $l$.