It is well known that $e^x$ can be expressed as a limit of a sequence:
$$e^x=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.$$
I can't help but to notice that the terms in parenthesis are just the first two terms of the Taylor series for $e^{x/n}$ centered at $0$:
$$e^{x/n}=\sum_{i=0}^\infty \frac{x^i}{n^ii!}.$$
Therefore, I conjecture the following:
Conjecture: Fix $x\in\mathbb R$. For each $N\in\mathbb N$, we have $$\lim_{n\to\infty}\left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^n=e^x.$$
I have verified this conjecture for various $N$ so I think it is true. I am now trying to figure out the formal proof of this conjecture.
Intuitively, the reason this conjecture is true is that as $n$ gets bigger, the sum will be quite close to $e^{x/n}$ and $\left(e^{x/n}\right)^n=e^x$. I need to formalize this idea. The absolute value difference between $\left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^n$ and $e^x$ is:
$$\left|e^x - \left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^n\right|=\left|\left(e^{x/n}\right)^n - \left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^n\right|=\left|\left(\sum_{i=0}^\infty\frac{x^i}{n^ii!}\right)^n - \left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^n\right|=\left|\sum_{i=N+1}^\infty\frac{x^i}{n^ii!}\right|\left|\sum_{k=0}^{N-1}\left(\sum_{i=0}^\infty\frac{x^i}{n^ii!}\right)^{N-1-k}\left(\sum_{i=0}^N\frac{x^i}{n^ii!}\right)^k\right|.$$
I am a bit lost on how to bound the second term which involves nested sums.