Is a monoid finitely generated by finite order elements periodic? This may be a simple question, but I can't make it out right now.

Let $M$ be a monoid, and $g_1, \ldots, g_n \in M$ be elements of finite order in $M$.  Is $\langle g_1, \ldots, g_n \rangle$ periodic (i.e, have all its elements of finite order)?

The monoid I'm  interested in is the multiplicative monoid of $n \times n$ matrices over $\mathbb{N}$.
Thanks.
 A: (Edit because I misread the question somewhat.)
To show that this is false for an arbitrary monoid it suffices to exhibit a monoid generated by elements of finite order which has elements of infinite order.  Take, for example, $M = \langle a, b | a^2 = b^2 = 1 \rangle$.  In this monoid $ab$ has infinite order.  One can see this geometrically by constructing the following linear representation: let $a$ and $b$ act by reflection across two lines through the origin at an angle $\theta$ with respect to each other which is not a rational multiple of $\pi$.  (For this reason $M$ is known as the infinite dihedral group.)
Probably the most important example of this phenomenon "in nature" is the modular group, which has group presentation $\langle a, b | a^2 = b^3 = 1 \rangle$ and hence is generated by an element of order $2$ and an element of order $3$.
I am not sure about your particular $M$.  Does $\mathbb{N}$ contain $0$ for you?  (If not I am having trouble finding any nontrivial elements of finite order.)
A: For the case you're interested in it is true, because the set of finite order $n$-by-$n$ matrices over $\mathbb{N}$ is precisely the set of permutation matrices.  In fact, every invertible element of $M_n(\mathbb{N})$ is a permutation matrix.  (Here $\mathbb{N}$ contains $0$, and invertible means invertible in the monoid $M_n(\mathbb{N})$ with multiplication.)
Suppose that $B=(b_{ij})\in M_n(\mathbb{N})$ is not a permutation matrix.  Then either one of the entries of $B$ is greater than $1$, one of the rows or columns of $B$ has more than one nonzero entry, or one of the rows or columns of $B$ is zero.   Let $A=(a_{ij})\in M_n(\mathbb{N})$ be invertible.
Suppose first that $B$ has an entry greater than $1$, i.e., $b_{i_0j_0}\gt1$ for some $i_0$ and $j_0$.  Since $A$ is invertible, its $i_0^\text{th}$ column is not zero, so there is an $i_1$ such that $a_{i_1i_0}\gt0$.  Then if $AB=(c_{ij})$, we have $c_{i_1j_0}\geq a_{i_1i_0}b_{i_0j_0}\gt1$, so that $AB\neq I$.  Hence, $B$ is not invertible in this case.  
Now suppose that $B$ has a row with more than one nonzero entry, i.e., $b_{i_0j_0}\gt0$ and $b_{i_0j_1}\gt0$ for some $i_0$ and $j_0\neq j_1$.  Again, there is an $i_1$ such that $a_{i_1i_0}\gt0$, and we have $c_{i_1j_0}\geq a_{i_1i_0}b_{i_0j_0}\gt0$ and $c_{i_1j_1}\geq a_{i_1i_0}b_{i_0j_1}\gt0$, so again $AB\neq I$.  A similar argument applies if $B$ has a column with more than one nonzero entry, either by multiplying in the other order, or by considering the transpose.  Thus in either case $B$ is not invertible.
This leaves only the case that $B$ has a zero row or column, in which case it is even a zero divisor.  Thus in any case, $B$ is not invertible.
