Show that $S\subset \mathbb{N}$ generates the group $\mathbb{Z}$ iff $n_1s_1 + \dots + n_ks_k =1$ 
Show that a subset $S$ of $\mathbb{N}$ generates the group $\mathbb{Z}$ of all integers iff there exist
$s_1, ..., s_k$ in $S$ and $n_1, ...,n_k$ in $\mathbb{Z}$ such that $n_1s_1 + \dots + n_ks_k =1$.

It can be proved that if $S$ is a nonempty subset of a group $G$ then
$\langle S\rangle=\{s_1^{n_1}s_2^{n_2}\cdots s_k^{n_k}\mid s_i\in S\text{ for }1\leq i\leq k,n_1,\cdots,n_k\in \mathbb{Z}\}$
where $\langle S\rangle=\cap_{H\in F}\quad\forall\quad \{H \mid H\leq G,S\subset H\}$ which is the smallest subgroup of $G$ containing $S$.
i.e., If $S$ is a subset of a group $G$ then $\langle S\rangle$ is the subgroup generated by $S$ (every element of $\langle S\rangle$ can be expressed as a combination under group operation of finitely many elements of the subset $S$).
If the group binary operation is addition then any element is of the form $n_1s_1+\cdots+n_ks_k$
My Attempt
A subset $S$ of $\mathbb{N}$ generates the group $\mathbb{Z}$ implies, for any $m\in \mathbb{Z}$ we can write
$m=m_1s_1 + ... + m_ks_k$ where $s_i\in S$ and $m_i\in \mathbb{Z}$
But where does the condition $n_1s_1 + ... + n_ks_k =1$ comes in ?

Once $1\in \langle S\rangle$, we can be sure that $S$ generates $\Bbb Z$. Can you see why?
Also, since $1\in\Bbb Z$, if $S$ generates $\Bbb Z$, then there must be a linear combination of elements in $S$ that is equal to $1$.

Your confusion might be because you're viewing $\langle S\rangle$ multiplicatively, not additively.
