I can't seem to prove that this is true, and in fact taking various examples with i roughly half of n and n growing, the term gets smaller as n gets larger.

Why am I confused? No, correct that. I'm usally confused. What I mean is, where's the error in my logic or calculation?

That is correct.

>And the ith term of this series is 1/(i-1)!

You mean the ith term of e^1, right?

>But the ith term is also
> n(n-1)(n-2)(n-3)...(n-(i-2))/(n^(i-1))(i-1)!

"Also" is inappropriate here. At best you may claim that the limit of
n(n-1)(n-2)(n-3)...(n-(i-2))/(n^(i-1))(i-1)! as n goes to infinity is 1/(i-1)!.

I would accept that, yes.

>It would seem, then, that as n->infinity, the term
> n(n-1)(n-2)(n-3)...(n-(i-2))/n^(i-1) must approach 1,

Yes, I would expect that.

>I can't seem to prove that this is true,

lim (n - a)/n = 1

for any a. The limit of a finite product of such ratios is also 1:

lim (n - a)/n · (n - b)/n · ... · (n - c)/n = 1,

for given (and independent of n) a, b, ..., c.

>and in fact taking various examples with
>i roughly half of n and n growing, the
>term gets smaller as n gets larger.

You can't take i "roughly half of n".
i must be fixed as n grows.

>Why am I confused? No, correct that. I'm usally confused.
>What I mean is, where's the error in my logic or
>calculation?

If i is "roughly half of n" then obviously the ith term must tend to 0. For, otherwise, the sum of an infinite number of such terms would not be finite.