How to proof that for all integers n>0 : the summation of i! from i=1 to n is odd? I could copy the right symbol on the title so here is a picture of it: https://imgur.com/a/Gd6sd. I looked for the lemma's because i didnt know what a lemma was and i totally didnt know what the lemma involving the parity of added numbers and that are multiplied was, but i couldt find anything that made me wiser
 A: $x!$ is even if $x>2 $ (follows from the fact that we would have a factor of 2 in $x!$)
$\sum_1^nx! = 1! +2!+3! ....x!$ since all $i!$ are divisible by for $i>1$
$\sum_1^nx! = 1! +0+0 ....0 \ mod \  (2)$
$\sum_1^nx! = 1 + 0 \ (mod \ 2)$
A: A "lemma" is just a minor result you prove along the way to your main result - think of it as a "mini-proof".
The statement of the hint is pretty cryptic. The idea is that addition and multiplication do particular things to even and odd numbers - things you already know, from basic arithmetic. In particular, the first lemma could be phrased as "The sum of an odd number and an even number is still odd"; the second could be "an even number times any number is still even".
Now, using our second lemma we can prove that $n!$ is even whenever $n \geq 2$: $2!$ is even, and $n!$ is just $2!$ times a bunch more numbers.
Using our first lemma, we can finish the proof: $\sum_1^ni! = 1! + 2! + \cdots + n!$. $2!$, $3!$, and so on up to $n!$ are all even, by what we just proved; so this is $1!$ plus a bunch of even numbers. Since $1!$ is odd, our first lemma tells us that this sum must also be odd. This concludes our proof.
Now, this was a messy way to put it - I phrased this proof to make it easier to understand, but that means it's not written in a mathematically precise way. To make it precise, you should re-phrase it using mathematical induction.
