For $pin (0, 1)$ and $nrightarrow infty$, how do I evaluate a general term in the binomial expansion of $(p+1-p)^n$?

I think not using the Binomial Theorem is making things very unnecessarily complicated.

From Binomial Theorem, we have

$${(a+b)^n = sum_{i=0}^{n}{nchoose i}a^ib^{n-i}}$$

Now, take ${1^n}$ as you did. Now take some non-zero number $p$. Any non-zero number. It does not have to be between $0$ and $1$. Then it's always true that

$${1^n = (p + (1-p))^n}$$

And now using Binomial Theorem with ${a=p}$, ${b=(1-p)}$ you get

$${(p+(1-p))^n = sum_{i=0}^{n}{nchoose i}p^i(1-p)^{n-i}}$$

But ${(p+(1-p))^n = 1^n = 1}$. So the sum always ${=1}$, always!

$${sum_{i=0}^{n}{nchoose i}p^i(1-p)^{n-i}=1}$$

It does not matter if ${n}$ was ${1}$, ${2}$, ${10}$, ${500}$, ${1000^2}$... the sum is always $1$. And there is nothing strange about this. If you put larger and larger values of $n$ - I agree, each term will get smaller and smaller. However, you are adding more and more terms to compensate. To really hammer this home...

$${frac{1}{2}=frac{1}{2}}$$

But also

$${frac{1}{4} + frac{1}{4} = frac{1}{2}}$$

But also

$${frac{1}{8} + frac{1}{8} + frac{1}{8} + frac{1}{8} = frac{1}{2}}$$

But also

$${frac{1}{16} + frac{1}{16} + frac{1}{16} + frac{1}{16} + frac{1}{16} + frac{1}{16} + frac{1}{16} + frac{1}{16}=frac{1}{2}}$$

... And so on. If I keep doing this, each term is becoming much, much smaller. But it always sums to ${frac{1}{2}}$, by construction. There are more and more terms to compensate for the fact that each term is smaller. This also helps to motivate the fact that the expression

$${0times infty}$$

is an indeterminate form. We cannot assign any value to ${0times infty}$ - since the answer depends on context. So indeed for any real number ${pneq 0}$, you have that

$${lim_{nto infty}sum_{i=0}^{n}{nchoose i}p^i(1-p)^{n-i}=1neq 0}$$

and there isn't anything strange about this result.

Each term will go to zero but you have more and more terms as $n to infty$.