$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.

\begin{align}
f(x) &=ax^6 +bx^5+cx^4+dx^3+ex^2+fx+3
\end{align}
Given that :
$f(1)= 1$, $f(2) =2$ , $f(3) = 3$, $f(4) =4$, $f(5)=5$, $f(6) =6$,
find $f(7)$.
It is also useful to add that $f(0)=3$.
Application of successive differences
lowers the degree of the polynomial on every step
down to $0$, and this useful formula shows up:
\begin{align}
\sum_{k=0}^7 (-1)^k\binom{7}{k}f(x-k)
&=0
\end{align}
\begin{align}
f(7)&=
-\sum_{k=1}^7 (-1)^k\binom{7}{k}f(7-k)
\\
&=
7 f(6)-21 f(5)+35 f(4)-35 f(3)+21 f(2)-7 f(1)+f(0)
\\
&=
7\cdot 6-21 \cdot 5+35 \cdot 4-35 \cdot 3 +21 \cdot 2 -7\cdot 1+3
\\
f(7)&=10.
\end{align}
