[proofplan]
Fix a nonnegative integer $k$ and use the binomial probability mass formula for all sufficiently large $n$. The [binomial coefficient](/page/Binomial%20Coefficient) factor separates into a finite product that tends to $1$, while the remaining power term separates into the standard exponential limit $(1-\lambda/n)^n \to e^{-\lambda}$ and a correction factor tending to $1$. Multiplying these limits gives the desired Poisson mass at $k$.
[/proofplan]
[step:Fix the mass level and write the binomial probability]
Fix $k \in \{0,1,2,\dots\}$. Since $n \to \infty$ and $k$ is fixed, there exists an integer $M \geq N_\lambda$ such that $n \geq k$ and $n > \lambda$ for every integer $n \geq M$. For such $n$, the parameter $\lambda/n$ belongs to $(0,1)$ and the binomial probability mass formula gives
\begin{align*}
\mathbb P(X_n = k)
=
\binom{n}{k}
\left(\frac{\lambda}{n}\right)^k
\left(1-\frac{\lambda}{n}\right)^{n-k}.
\end{align*}
Terms with $N_\lambda \leq n < M$ do not affect the limit as $n \to \infty$.
[guided]
We fix one value $k \in \{0,1,2,\dots\}$ because the theorem asks for pointwise convergence of the probability mass functions. The integer $k$ does not vary with $n$. Choose an integer $M \geq N_\lambda$ such that $M \geq k$ and $M > \lambda$. Then for every integer $n \geq M$, we have both $n \geq k$ and $0 < \lambda/n < 1$.
The condition $n \geq k$ ensures that the event $\{X_n = k\}$ lies in the support of the binomial law. The condition $0 < \lambda/n < 1$ ensures that the ordinary binomial mass formula has no endpoint ambiguity. Thus, for every integer $n \geq M$,
\begin{align*}
\mathbb P(X_n = k)
=
\binom{n}{k}
\left(\frac{\lambda}{n}\right)^k
\left(1-\frac{\lambda}{n}\right)^{n-k}.
\end{align*}
Since changing or discarding finitely many initial terms of a sequence does not change its limit, it is enough to compute the limit of this expression along integers $n \geq M$.
[/guided]
[/step]
[step:Separate the binomial coefficient factor into a finite product]
For $n \geq M$, define the finite product $A_n(k)$ by
\begin{align*}
A_n(k) := \prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right),
\end{align*}
with the convention that $A_n(0) := 1$. If $k = 0$, then
\begin{align*}
\binom{n}{0}\left(\frac{\lambda}{n}\right)^0 = 1 = \frac{\lambda^0}{0!}A_n(0).
\end{align*}
If $k \geq 1$, then
\begin{align*}
\binom{n}{k}\left(\frac{\lambda}{n}\right)^k
=
\frac{n(n-1)\cdots(n-k+1)}{k!}\frac{\lambda^k}{n^k}
=
\frac{\lambda^k}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)
=
\frac{\lambda^k}{k!}A_n(k).
\end{align*}
For each fixed index $j \in \{0,\dots,k-1\}$,
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{j}{n}\right)=1.
\end{align*}
Because the product defining $A_n(k)$ has finitely many factors, the finite product law for limits yields
\begin{align*}
\lim_{n \to \infty} A_n(k)=1.
\end{align*}
Therefore
\begin{align*}
\lim_{n \to \infty}\binom{n}{k}\left(\frac{\lambda}{n}\right)^k
=
\frac{\lambda^k}{k!}.
\end{align*}
[/step]
[step:Identify the limiting exponential factor]
For $n \geq M$, define
\begin{align*}
B_n(k) := \left(1-\frac{\lambda}{n}\right)^{n-k}.
\end{align*}
Since $1-\lambda/n \neq 0$ for $n \geq M$, we may write
\begin{align*}
B_n(k)
=
\left(1-\frac{\lambda}{n}\right)^n
\left(1-\frac{\lambda}{n}\right)^{-k}.
\end{align*}
The standard exponential limit gives
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^n = e^{-\lambda}.
\end{align*}
Also,
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^{-k} = 1,
\end{align*}
because $1-\lambda/n \to 1$ and the map $t \mapsto t^{-k}$ is continuous at $t=1$. Hence the finite product law for limits gives
\begin{align*}
\lim_{n \to \infty}B_n(k)=e^{-\lambda}.
\end{align*}
[guided]
The remaining factor is
\begin{align*}
B_n(k) := \left(1-\frac{\lambda}{n}\right)^{n-k}.
\end{align*}
The exponent $n-k$ is close to $n$, so we separate the expression into a main exponential part and a fixed correction. Since $n \geq M$ implies $n > \lambda$, the base $1-\lambda/n$ is positive and therefore nonzero. Thus the algebraic identity
\begin{align*}
\left(1-\frac{\lambda}{n}\right)^{n-k}
=
\left(1-\frac{\lambda}{n}\right)^n
\left(1-\frac{\lambda}{n}\right)^{-k}
\end{align*}
is valid.
Now we evaluate the two factors separately. The standard exponential limit states that, for fixed $\lambda > 0$,
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^n = e^{-\lambda}.
\end{align*}
The correction factor has fixed exponent $-k$. Since
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)=1,
\end{align*}
and the real function $t \mapsto t^{-k}$ is continuous at $t=1$, we obtain
\begin{align*}
\lim_{n \to \infty}\left(1-\frac{\lambda}{n}\right)^{-k}=1.
\end{align*}
Multiplying these two limits gives
\begin{align*}
\lim_{n \to \infty}B_n(k)=e^{-\lambda}.
\end{align*}
[/guided]
[/step]
[step:Multiply the factor limits to obtain the Poisson mass]
For every integer $n \geq M$,
\begin{align*}
\mathbb P(X_n = k)
=
\left[\binom{n}{k}\left(\frac{\lambda}{n}\right)^k\right]B_n(k).
\end{align*}
The first bracket tends to $\lambda^k/k!$, and $B_n(k)$ tends to $e^{-\lambda}$. By the finite product law for limits,
\begin{align*}
\lim_{n \to \infty}\mathbb P(X_n = k)
=
\frac{\lambda^k}{k!}e^{-\lambda}.
\end{align*}
Since $k \in \{0,1,2,\dots\}$ was arbitrary, the asserted convergence holds for every nonnegative integer $k$.
[/step]
