Writing a proof I am stuck at showing how if:
$P \rightarrow Q$ then this implies ($Q \rightarrow R) \rightarrow (P \rightarrow R)$
I know that if we assume $P$ is true then $Q$ also must be true. 
Therefore if $Q$ is true then $R$ must be true. And because $Q$ is true because $P$ is true so therefore $R$ must also be true.  
However it would be appreciated if someone could help me understand this why is the case and what proof technique it is. 
 A: You can also prove this somewhat verbosely with a truth table... (I have named my intermediate steps just because the table was too wide to display nicely.)
$$
\begin{array}{|c c c|c|c|c|c|c|}
P & Q & R & P \implies Q & Q \implies R & P \implies R &  & \\
 & & &A&B&C&B \implies C & A \implies (B \implies C) \\
\\
T & T & T & T & T & T & T & T\\
T & T & F & T & F & F & T & T\\
T & F & T & F & T & T & T & T\\
T & F & F & F & T & F & F & T\\
F & T & T & T & T & T & T & T\\
F & T & F & T & F & T & T & T\\
F & F & T & T & T & T & T & T\\
F & F & F & T & T & T & T & T\\
\end{array}
$$
A: This property is called transitivity of implication.
You pretty much gave a proof in your question, but here's a proof written out in slightly more precise terms.
Suppose $P \Rightarrow Q$ is true. To prove $(Q \Rightarrow R) \Rightarrow (P \Rightarrow R)$, you need to assume $Q \Rightarrow R$ and derive $P \Rightarrow R$. So assume that $Q \Rightarrow R$ is true. To prove $P \Rightarrow R$ is true, you need to assume $P$ is true and derive $R$. So assume $P$ is true. All we have to do now is prove that $R$ is true.
At this point, we're assuming that $P \Rightarrow Q$, $Q \Rightarrow R$ and $P$ are all true. So:


*

*Since $P$ and $P \Rightarrow Q$ are true, we have that $Q$ is true; and

*Since $Q$ and $Q \Rightarrow R$ are true, we have that $R$ is true.


So we're done.
A: If you want a conceptual proof, you just need to know how to prove an implication. How do you prove $A\rightarrow B$? Assume $A$ is true, then show that $B$ follows (hoping you have some other information laying around to do this using $A$.


*

*Step 1 To show:
$(P \rightarrow Q) \rightarrow ((Q \rightarrow R) \rightarrow (P \rightarrow R))$,
assume $(P \rightarrow Q)$ and show $(Q \rightarrow R) \rightarrow (P \rightarrow R)$.

*step 2 To show $(Q \rightarrow R) \rightarrow (P \rightarrow R)$, assume  $Q \rightarrow R$ and then show $P \rightarrow R$.

*step 3 To show $P \rightarrow R$, assume $P$ and then show $R$.
At each stage we're assuming something so by this last step we've assumed a whole bunch that can help us to show $R$.
Using $P$, assumed true from the last step and $P\rightarrow Q$ assumed true from step 1, $Q$ follows, via modus ponens.
Using this $Q$ and $Q\rightarrow R$ assumed true from step 2, $R$ follows, again via modus ponens.
And we're done. In short, a chain of implications unraveled from left to right.
A: It’s a sort of transitivity. $P\implies Q$ is true. So if $Q\implies R$, then $P\implies Q\implies R$. I don’t think there is a name for this. It’s just multiple implications that give a result. 
Proving this is simple. Given $Q\implies R$, we want to demonstrate $P\implies R$.
So let’s suppose that $P$ is true. So, $Q$ is true by $(P\implies Q)$ (it’s called the “Modus ponens”). But $Q$ is true, so $R$ is true. 
QED.
A: $A\Rightarrow B$ can be written as $\neg A\vee B.$ Therefore,
$$
(Q\Rightarrow R)\Rightarrow(P\Rightarrow R) \\
=\neg(\neg Q\vee R)\vee (\neg P\vee R) \\
= (Q \wedge \neg R)\vee (\neg P\vee R) \\
= (\neg P \vee Q\vee R)\wedge (\neg P \vee R \vee \neg R) \\
= (\neg P \vee Q\vee R) \\
= (\neg P \vee Q)\vee R \\
= (P\Rightarrow Q) \vee R
$$
