Solve the functional equation $f(xy) = f(x)f(y) - f(x + y) + 1$ and $f(1) = 2$ I'm working on this problem already, the domain and co-domain are the set of Rational Numbers. we see that setting $x = 1$ and $y = n$, we obtain $f(n + 1) = f(n) + 1$
But I'm trying to prove that $f(n) = n + 1$ for all $n$, the case when $n = 1, 2$ is obviously true, how do I approach this via induction? 
 A: 
The condition $f(1)=2$ is unnecessary.  The only functions $f:\mathbb{Q}\to\mathbb{C}$ sastisfying $$f(xy)=f(x)\,f(y)-f(x+y)+1\tag{$\star$}$$
  are the constant function $f\equiv 1$ and the function $f(x)=x+1$ for all $x\in\mathbb{Q}$.

First, as ajotaxe observed, $f(0)=1$.  If $f(1)=1$, then we obtain from $(\star)$ that
$$f(x)=f(x\cdot 1)=f(x)\,f(1)-f(x+1)+1=f(x)-f(x+1)+1$$
for all $x\in\mathbb{Q}$.  This means $f(x+1)=1$ for all $x\in\mathbb{Q}$, which leads to $f\equiv 1$.  From now on, we assume that $f(1)\neq 1$.  
By plugging in $x:=1$ and $y:=-1$ into $(\star)$, we  have
$$f(-1)=f(1)\,f(-1)\,.$$
Because $f(1)\neq 1$, we must have $f(-1)=0$.  Putting $y:=-1$ yields
$$f(-x)=-f(x-1)+1\,.\tag{*}$$
Now, plugging in $x:=1$ and $y:=-2$ into $(\star)$, we obtain
$$f(-2)=f(1)\,f(-2)-f(-1)+1=f(1)\,f(-2)+1\,.$$
Using $(*)$, we have $f(-2)=-f(1)+1$, so the equation above translates to
$$-f(1)+1=-\big(f(1)\big)^2+f(1)+1\,.$$
This gives
$$\big(f(1)\big)^2-2\,f(1)=0\,.$$
That is, $f(1)=0$ or $f(1)=2$.
If $f(1)=0$, then $y:=1$ into $(\star)$ leads to
$$f(x)=-f(x+1)+1\,.\tag{$\Box$}$$
This proves that $f(n)=\frac{1+(-1)^n}{2}$ for every integer $n$, and from $(\Box)$, we get
$$f(x+2)=f(x)$$
for all $x\in\mathbb{Q}$.  Now, if we plug in $y:=n$ into $(\star)$, where $n$ is an even integer, then
$$f(xn)=f(x)\,f(n)-f(x+n)+1=f(x)-f(x)+1=1$$
for every $x\in\mathbb{Q}$.  Thus, substituting $x:=\frac{1}{2}$ and $n:=2$ in the previous equation, we obtain $$0=f(1)=f\left(\frac{1}{2}\cdot 2\right)=1\,,$$
which is absurd.  Hence, $f(1)\neq 0$, and so $f(1)=2$ as required.
Now, $(\star)$ with $y:=1$ implies that
$$f(x+1)=f(x)+1$$
for all $x\in\mathbb{Q}$.  Ergo, we can see that $f(n)=n+1$ for every integer $n$.  Consequently, for a rational number $x=\frac{p}{q}$, where $p,q\in\mathbb{Z}$ with $q>0$, we have
$$p+1=f(p)=f\left(\frac{p}{q}\cdot q\right)=f\left(\frac{p}{q}\right)\,f(q)-f\left(\frac{p}{q}+q\right)+1=(q+1)\,f\left(\frac{p}{q}\right)-f\left(\frac{p}{q}\right)-q+1\,.$$
Hence,
$$f(x)=f\left(\frac{p}{q}\right)=\frac{p+q}{q}=\frac{p}{q}+1=x+1\,,$$
as desired.
P.S.: It is a very interesting question what happens if the domain extends to $\mathbb{R}$.  All I know is that, if a nonconstant function $f:\mathbb{R}\to\mathbb{Q}$ satisfies $(\star)$, then
$$f(x+r)=f(x)+r$$
and
$$f(rx)=r\,f(x)-r+1$$
for all $x\in\mathbb{R}$ and $r\in\mathbb{Q}$.
A: For $x=y=0$,
$$f(0)=f(0)^2-f(0)+1$$
Then $f(0)=1$.
With the equation $f(n+1)=f(n)+1$, you have it (see Thomas Andrews' comment).
