I have proved that for $k\ge 4$ the sum $$\sum_{i=1}^k i!$$ can never be a perfect square i.e., of the form $n^2\, (m=2).$ But am struggling with the generalisation i.e., with the form $n^m$. Is the statement valid? I found this question on quora and no one has answered it.. Please suggest something... Thanks in advance.
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Since the possible values for the unit digit of a perfect square number are $0, 1,4,5,6$ and $9.$ Why? Take a number $pq,$ where $0\le q\le 9.$ Note that $pq$ could be any digit number. Now $(pq)^2=(10p+q)^2=10(10p^2+2pq)+q^2.$ So unit digit of $(pq)^2$ and $q^2$ is basically the same. As $0\le q\le 9$ so unit digit of $q^2$ would be one among $0, 1,4,9,6,5.$ Now $\sum_{i=1}^{4}i!=33.$ And for $k\ge 5,$ $k!$ contains $5!$. Unit digit of $5!$ is $0.$ Hence unit digit of $\sum_{i=5}^k i! $ is also $0.$ This implies unit digit of $\sum_{i=1}^k i!=\sum_{i=1}^ 4 i!+\sum_{i=5}^k i!$ is $3+0=3.$ – Dhrubajyoti Bhattacharjee May 27 '20 at 14:20
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(Continue...) But since a perfect square number can never have $3$ as a unit digit. So $\sum_{i=1}^k i!$ can never be a perfect square for $k\ge 4.$ Although for $k\le 3,$ specifically for $k=1$ and $k=3$ the sum $\sum_{i=1}^ k i!$ yields a perfect square. – Dhrubajyoti Bhattacharjee May 27 '20 at 14:24