The homomorphism theorem says that from a surjective homomorphism $\phi\colon R\to S$ we can define an isomorphism
$$
\tilde{\phi}\colon R/\ker\phi\to S
$$
by mapping each coset $r+\ker\phi$ to $\phi(r)$.
If $x,y\in \ker\phi$, you have to show that
$$
(1+x)(1+y)\in 1+\ker\phi
$$
and that every element in $1+\ker\phi$ has an inverse which still belongs to $1+\ker\phi$. This ensures $1+\ker\phi$ is a subgroup of $R^*$.
For normality, you have to show that, for $x\in\ker\phi$ and $r\in R^*$, $r(1+x)r^{-1}\in 1+\ker\phi$.
Next, you have to show that $\phi^*\colon R^*\to S^*$ (the restriction of $\phi$) induces a morphism $R^*/N\to S^*$, that is, $N\subseteq \ker\phi^*$.
Just for completeness, here's a sketch.
If $x,y\in\ker\phi$, then $(1+x)(1+y)=1+(x+y+xy)\in 1+\ker\phi=N$, because $x+y+xy\in\ker\phi$.
If $x\in\ker\phi$, set $y=-x$; then we know that $y^n=0$, for some $n>0$. Then
$$(1-y)(1+y+y^2+\dots+y^{n-1})=1-y^n=1$$
and so $1-y=1+x$ is invertible (the element is obviously also a left inverse).
If $x\in\ker\phi$ and $r\in R^*$, then $r(1+x)r^{-1}=1+rxr^{-1}\in N$.
Now $\phi$ induces a group morphism $\phi^*\colon R^*\to S^*$, because invertibles in $R$ are mapped to invertible elements of $S$ (because $\phi$ is surjective). It is obvious that $N\subset\ker\phi^*$, because $\phi^*(1+x)=\phi(1+x)=1+\phi(x)=1$.
Conversely, if $r\in\ker\phi^*$, then $\phi(r)=1$, so $r-1\in\ker\phi$ and $r=1+(r-1)\in N$.
I think it is better to use the adjunction relation from exercise 1.8, which tells us that it is equivalent to prove surjectivity of the homomorphism $\mathcal{O}_{Proj(S),\varphi^{-1}(p)}\to\mathcal{O}_{Proj(T),p}$ (using that $f^{-1}G_x=G_{f(x)}$ for a sheaf $G$). Then it is possible to apply proposition 2.5a), which gives us the homomorphism $S_{(\varphi^{-1}(p))}\to T_{(p)}$.
Best Answer
First reduce to the case where $n$ and $m$ are both powers of a prime $p$, using the Chinese remainder theorem. After this reduction the result is clear since the $p$-adic valuation of $n$ is lesser than the $p$-adic valuation of $m$ for any prime $p$.