[bigpooch]

2) There are several methods, but maybe I am too old to
think of what is considered the simplest proof.

1) Too easy:
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For n&gt;3, the last digit of the sum ends in 3, so obviously
can't be a square.
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Better problem IMHO is to show that 1! + 2! +...+ n! is not
a kth power for n&gt;3, k&gt;=2.
Hint for this:
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note the sum is divisible by 9 for n&gt;=5;
for n&gt;=8, look at (mod 27)
and simply check for n&lt;8
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