$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.

Since $f(x) - x$ is a polynomial of degree $6$ and has $6$ roots $1, 2, 3, 4, 5, 6$ by condition, we can factorize $f(x) - x$ as:
$$f(x) - x = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5)(x - 6).$$
Plug in $x = 0$ in the above expression, we have
$3 - 0 = C\times 6!$, hence $C = \dfrac{3}{6!}$.
Therefore, 
$$f(7) = 7 + (f(7) - 7) = 7 + \frac{3}{6!}(7 - 1)(7 - 2)(7 - 3)(7 - 4)(7 -5 )(7 - 6) = 7 + \frac{3}{6!}\times 6! = 10.$$
