Hartshorne's Exercise II.5.1 - Projection formula I'm trying to solve Exercise 5.1 of Chapter II of Hartshorne - Algebraic Geometry.
I'm fine with the first $3$ parts, but I'm having troubles with the very last part, which asks to prove the projection formula:

Let $f:X\to Y$ be a morphism of ringed spaces, $\mathscr{F}$ an $\mathcal{O}_X$-module and $\mathcal{E}$ a locally free $\mathcal{O}_Y$-module of finite rank. Then there is a natural isomorphism
  $$ f_*(\mathscr{F}\otimes f^*\mathcal{E}) \;\cong\; f_*(\mathscr{F})\otimes \mathcal{E} $$

After thinking quite a long time about it, I checked on the internet and I found the following solution:

$$
\begin{eqnarray} 
f_*(\mathscr{F}\otimes f^*\mathcal{E}) 
&\;\cong\;& f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n}) \\\\
&\;\cong\;& f_*(\mathscr{F}\otimes \mathcal{O}_X)^{n} \\\\ 
&\;\cong\;& f_*(\mathscr{F})^{n} \\\\ 
&\;\cong\;& f_*(\mathscr{F})\otimes \mathcal{O}_Y^{\,n} \\\\ 
&\;\cong\;& f_*(\mathscr{F})\otimes \mathcal{E} \\\\ 
\end{eqnarray}
$$

Is this correct? If it is, could you explain me why do we have the isomorphism
$$ f_*(\mathscr{F}\otimes \mathcal{O}_X^{\,n})
\;\cong\; f_*(\mathscr{F}\otimes \mathcal{O}_X)^{n} \quad ?  $$

The adjunction formula in Hartshorne, and the one I proved myself, is that 
$$ \hom_X(f^*E,F) \simeq \hom_Y(E,f_*F)$$
How does the one used here follow from this one? Is it general non-sense? 
