Find all natural numbers such that $\sum_{k=1}^{n} \frac{n^k}{k!}$ is an integer.
I've tried to bring all fractions under commmon denominator and it didn't helped me much. With guessing I find out that only $n=1,2,3$ satisfy the condition, but I can't prove that they are the only ones.
I tried to evaluate the series and then work with the result, but I didn't made any progress.
Clearly $n=1$ works. Consider $n \geq 2$. $\sum_{k=1}^{n}{\frac{n^k}{k!}}$ is an integer, so $(n-2)!\sum_{k=1}^{n}{\frac{n^k}{k!}}$ is an integer.
Since $(n-2)!\sum_{k=1}^{n-2}{\frac{n^k}{k!}}$ is an integer, we have that $(n-2)!\sum_{k=n-1}^{n}{\frac{n^k}{k!}}$ is an integer, i.e. that $\frac{n^{n-1}}{n-1}+\frac{n^n}{(n-1)n}=\frac{2n^{n-1}}{n-1}$ is an integer.
Since $(n-1, n)=1$, we have that $n-1 \mid 2$, so $n=2, 3$.