Prove that $n^k<(1+a)^n$ for sufficiently large $n$

Question

Prove using the binomial theorem, that for natural numbers $n$ and $k$ and real positive number $a$, $n^k<(1+a)^n$ for all $n>N$.

Using the binomial theorem, $(1+a)^n=1+na+\dots+\frac{n(n-1)\dots(n-k+1)(n-k)}{(k+1)!}a^{k+1}+\dots+a^n$

For sufficiently large $n$, $\frac{n(n-1)\dots(n-k+1)(n-k)}{(k+1)!}a^{k+1}\sim\frac{n^k(n-k)}{(k+1)!}a^{k+1}$

So the inequality is equivalent to the fact $\frac{n-k}{(k+1)!}a^{k+1}>1$ for sufficiently large $n$, which can be made true by making $n>\frac{(k+1)!}{a^{k+1}}+k$

Is this method of reasoning valid, do I need to be more careful with approximating things in the limit and is there a better way of establishing the inequality still using the binomial theorem (I know $\lim_{n \to \infty} n^{\frac{1}{n}}=1$ which also proves the inequality).

Answer 1

Suppose $n>k$, then $(1+a)^n = \sum_{i=0}^n \binom{n}{i} a^i \ge \binom{n}{k+1} a^{k+1}$.

$\binom{n}{k+1} = {n (n-1) \cdots (n-k) \over k!}\ge {(n-k)^{k+1} \over (k+1)!}$.

Then ${1 \over n^k} (1+a)^n \ge (n-k) (1-{k \over n})^k a^{k+1}$.

Since $\lim_{n \to \infty} (1-{k \over n})^k = 1$, we see for some $N$ we have ${1 \over n^k} (1+a)^n >1$ for $n \ge N$.

Answer 2

The structure of the argument is good. There is some imprecision connected with the $\sim$ part that should be fixed.

Let $n\gt 2k$. Then $n-1$, $n-2$, and so on up to $n-k$ are greater than $n/2$. It follows that $$(1+a)^n\gt n^k \frac{n}{2^k(k+1)!}a^{k+1},$$ and now we can specify precisely an $N$ that will work.