It is clear that
$$ \sum_{\vert \alpha \vert=k}\frac{k!}{\alpha!}=n^k,$$
where $k$ is a given positive integer and $\alpha\in \mathbb{N}^{n}$ such that $\alpha=(a_1, \dots, \alpha_n)$, $\vert \alpha \vert=\sum_{i=1}^{n}\alpha_i$ and $\alpha! = \alpha_1! \alpha_2! \cdots \alpha_n!$.
I would like to get an upper bound for the variant of the above sum with the additional restriction that $\alpha_i\ge 1$ for each $i$. I think I saw somewhere an estimate
$$\sum_{\vert \alpha \vert=k, \alpha_i\ge 1}\frac{k!}{\alpha!}\le \frac{n^k}{n!}.$$
Is the above estimate true and if so, could you help me prove it?
Your $\sum\limits_{\vert \alpha \vert=k}\frac{k!}{\alpha!}=n^k$ is counting the number of was of putting $k$ labelled balls into $n$ labelled boxes. $\alpha_i$ is the number of balls in the $i$th box.
I think your $\sum\limits_{\vert \alpha \vert=k, \alpha_i\ge 1}\frac{k!}{\alpha!} =n!\, S_2(k,n)$ using Stirling numbers of the second kind, since you are counting the number of ways of putting $k$ labelled balls into $n$ labelled boxes so each box has at least one ball.
Let's take an easy counter-example to your conjecture where $n=k=3$: here $\sum\limits_{\vert \alpha \vert=k, \alpha_i\ge 1}\frac{k!}{\alpha!} = \frac{3!}{1!1!1!}=6$ while $\frac{n^k}{n!} = \frac{3^3}{3!}=4.5$ which is smaller.