Proving all sufficiently large integers can be written in the form $ax+by$ 
Let $a,b \in \mathbb N \setminus \{0,1\}$ such that $\gcd(a,b)=1$
Let $F=\{ax+by \mid (x,y) \in \mathbb N^2\}$
Prove that all integers $\geq (a-1)(b-1)$ are in $F$, but that $(a-1)(b-1)-1\notin F$

Here are my thoughts :
I've noticed that $F$ is closed under addition and multiplication.
I also proved by contradiction that $(a-1)(b-1)-1\notin F$.
It is clear as well that $F$ is unbounded.
However, I can't even manage to prove that $(a-1)(b-1) \in F$ (Bézout fails), let alone higher integers.
 A: The numbers $a$ and $b$ are coprime so by the euclidean algorithm every integer can be written in the form $ka+sb$. Notice that $0=(b)a+(-a)b$. So every element can be written in the form $ka+sb$ with $0\leq k <b$.
Prove that every number can be written uniquely in this form. Prove that given a number if it can be written as $ka+sb$ with $k$ and $s$ non-negative then at can be written with $k$ and $s$ non-negative and with $0\leq k<b$.
So a number cannot be written with $k$ non-negative and $s$ non-negative if and only if it is of the form $ka+sb$ with $0\leq k<b$ and $s$ negative. Which is the largest number of this form?
Note this problem is known as the frobenius coin problem and the Chicken McNugget Theorem.
A: Hint: For an $n\geq (a-1)(b-1)$, show that one of the numbers $n-ka$ for $k\in \{0,1,\ldots,(b-1)\}$ is divisble by $b$.
A little more: It's quite obvious that the difference between any two numbers in that set is divisible by $a$, use $\gcd(a,b)=1$ to show that the remainder on division by $b$ take on all possible values.
