$f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h $, find $f(7)$. Problem : 

Given a polynomial $f(x) =ax^6 +bx^5+cx^4+dx^3+ex^2+gx+h$
  such that $f(1)= 1, f(2) =2 , f(3) = 3, f(4) =4, f(5)=5, f(6) =6$.
Find $f(7)$ in terms of $h$.

My approach: 
We can put the values of $f(1) = 1$ in the given equation and $f(2) = 2$, etc. But this is quite time consuming by making six different equations and then solve them to get the values of $a,b,c,d,e,g,h$.  Please suggest some alternate solution for this.

let $g(x) = f(x) - x$
Then:
$$g(1)=g(2)=g(3)=g(4)=g(5)=g(6)=0$$
So:
$$g(x) = A(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$$
We also know that $g(0)=f(0)-0=f(0)=h$ so we can find $A$.
$h=A\times 1\times 2\times 3\times 4\times 5\times 6$ so $A=\frac{h}{6!}$
$f(7) = g(7) + 7 = 6\times 5\times 4\times 3\times 2\times 1\times \frac{h}{6!} + 7 = h + 7$
