Monotonicity, boundaries and convergence of the sequence $ \left\{ \frac{a^n}{n!} \right\} $. everyone. 
I have a doubt on the following question:
Let $ \left\{ \frac{a^n}{n!} \right\}, n \in \mathbb{N} $ be a sequence of real numbers, where $ a $ is a positive real number.
a) For what values of $ a $ is the sequence above monotonous? And bounded?
b) For what values of $ a $ is the sequence above convergent? Determine the limit of the sequence on that case.
How do I start this question?
Clearly 0 is a lower boundary. But is there a $ n_{max} $ after which $ a^n $ is always smaller than $ n! $ ? 
As far as the monotonicity goes, I thought about splitting the problem into three cases:
$ \bullet $ For $ 0 < a < 1 $:
In this case, knowing that if $ 0 < a < 1 $, then $ a^{n+1} < a^n $, we can assume that $ A_{n+1} $ is always smaller than $ A_n $ because the numerator is decreasing (as showed) and the denominator is obviously increasing. Thus, if $ a_{n+1} < a_n, \forall \,\, n \in \mathbb{N} $, the sequence is strictly decreasing and so it is monotonic.
$ \bullet $ For $ a = 1 $:
In this case, the sequence is $ \frac{1}{n!} $ which is always positive, has 0 as a lower boundary and 1 as an upper boundary and thus it is bounded. Also, given that in this case $ A_{n+1} \leq A_n, \forall \,\, n \in \mathbb{N} $ the sequence is also monotonic.
$ \bullet $ For $ a > 1 $:
In this case, $ a^{n+1} > a^n, \forall \,\, n \in \mathbb{N} $. However, in order to known whether $ a_{n+1} < a_n $ or $ a_{n+1} > a_n $ we have to check if $ a^n > n! $, which I do not know how to do.
Could anyone help me with this question?

Thanks for the attention.
Kind regards,
Pedro

Hint :
$$\frac{a^n}{n!} = \frac{\underbrace{a \cdot a \cdot a... \cdot a}_{n \text{ times}}}{1\cdot 2 \cdot 3...\cdot n}$$
So for any finite positive parameter $a$ one can assess there are two numbers $n_1$ and $n_2$ such that $n_1 < a < n_2$, where both $n_1$, $n_2 \leq n$ as $n \rightarrow \infty$.  
What can one conclude henceforth? 
