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Is this following proof correct?

The limit \begin{align*} \lim_{n\to\infty}\frac{t^{n+1}}{(n+1)!} \end{align*} tends to zero if for every $\epsilon>0$, there exists $N_\epsilon$ such that \begin{align*} n\geq N_\epsilon\implies \left|\frac{t^{n+1}}{(n+1)!}\right|<\epsilon \end{align*} For any $t\in\mathbb{R}$, there exists $N\in\mathbb{N}$ such that $t<N$. Let $N$ be the least such natural number. Then
\begin{align*} \frac{t^{N+k}}{(N+k)!}&=\frac{t\cdot t\cdots t\cdot t\cdots t\cdot t}{1\cdot 2\cdots (N-1)N\cdots(N+k-1)(N+k)}\\ &<\frac{t\cdot t\cdots t \cdot t\cdots t\cdot t}{1\cdot 1\cdots 1\cdot N\cdots N\cdot N}\\ &=t^N\left(\frac{t}{N}\right)^{k} \end{align*} Let $a=\frac{t}{N}$. Since $t<N$, $a=\frac{t}{N}<1$. If \begin{align*} \frac{t^{N+k}}{(N+k)!}<t^N\left(\frac{t}{N}\right)^{k}=t^Na^{k}<\epsilon \end{align*} then dividing both sides by $t^N$, \begin{align*} a^{k}<\frac{\epsilon}{t^N} \end{align*} Taking $\log_a$ of both sides reverses the inequality because it is a decreasing function. Hence \begin{align*} k>\log_{a}\left(\frac{\epsilon}{t^N}\right). \end{align*} Any value for $k$ greater than this expression will ensure the limit remains within the desired bounds. Therefore, take $N_\epsilon=N+k$. Since the choice of $t$ was arbitrary, we may conclude that the limit converges throughout the entirety of $\mathbb{R}$.

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The proof looks good (actually very good) to me. The only improvement I can suggest is to turn it around a little, so that no cleverness is invoked.

Proof: Given $t>0$ and $\epsilon > 0$ arbitrary, put: $$\begin{align*} N &= \lfloor t \rfloor + 1 & (\text{$N$ is the smallest natural number strictly greater than 1}) \\ a &= \tfrac{t}{N} & (\text{so $a < 1$}) \\ k &= \left\lceil \log_{a}\left( \frac{\epsilon}{t^N} \right) \right\rceil \\ N_{\epsilon} &= N+k \end{align*}$$ Then if $n > N_{\epsilon}$ we have $$\begin{align*} \frac{t^{n+1}}{(n+1)!} &= \frac{t^{N+n-N+1}}{N!(N+1)(N+2)\dots(n+1)} \\ &< \frac{t^N}{1^N} \cdot \left( \frac{t}{N} \right)^{n+1-N} \\ &= t^N a^{n - N_{\epsilon} + N_{\epsilon} +1-N} \\ &< t^N a^{k+1} & (\text{via $N_{\epsilon} = N+k$, $n > N_{\epsilon}, a < 1$)} \\ &< t^N \frac{\epsilon}{t^N} & (\text{via construction of $k$}) \\ &= \epsilon \end{align*}$$ as required.

Remark: Your calculations did all the hard work of finding $N_{\epsilon}$. All this answer suggests is to summarize it into easier(?) steps from 'if $n > N_{\epsilon}$' to 'then $(\ldots) < \epsilon$'.

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