I got the following sum with which I want to prove one limit fact:


$$
f\_n(a) = \sum\limits\_{t=0}^{n-1} \binom{n-1}{t} (a^t)^{n-t}
$$


I want to prove that $f\_n(a) \to 1$ while $n \to \infty$ for $a\in [0,1)$ if true. (I plotted and it looks like it were true). As you see this is kind of binomial sum, but instead of multiplying by $a^{n-t}$ we are powering. I wrote the sum without sigma sign and tried to prove using members, however I stuck because the size of the sum still growing.


Any help would be appreciated.


Thank you.


*[edit]* If it is possible the speed of decreasing is also interesting, in terms of small-o.



Almost binomial sum limit
Splitting your sum as $\sum\_{t=0}^{n-1}=1+\sum\_{t=1}^{n-1}$, it suffices to show that $\sum\_{t=1}^{n-1}\binom{n-1}ta^{t(n-t)}\rightarrow0$ as $n\rightarrow\infty$; provided $0<a<1$.


To illustrate our proposed method, assume $0<a<\frac12$. Since $t(n-t)\leq n-1$, we estimate
$$0\leq\sum\_{t=1}^{n-1}\binom{n-1}ta^{t(n-t)}\leq a^{n-1}\sum\_{t=1}^{n-1}\binom{n-1}t\leq (2a)^{n-1}\rightarrow0, \qquad \text{as $n\rightarrow\infty$}.$$
The claim holds for $0<a<\frac12$. The general idea is quite similar.


Fix $k\in\mathbb{N}$ and assume $0<a<\sqrt[k]{\frac12}$. For $n\gg k$, "shave off" terms from above and below:
$$\sum\_{t=1}^{n-1}\binom{n-1}ta^{t(n-t)}
=\left(\sum\_{t=1}^{k-1}+\sum\_{t=k}^{n-k}+\sum\_{t=n-k+1}^{n-1}\right)
\binom{n-1}ta^{t(n-t)}$$
$$\qquad \qquad =\sum\_{t=1}^{k-1}\left[\binom{n-1}t+\binom{n-1}{t-1}\right]a^{t(n-t)}+
\sum\_{t=k}^{n-k}\binom{n-1}ta^{t(n-t)}$$
$$=\sum\_{t=1}^{k-1}\binom{n}ta^{t(n-t)}+
\sum\_{t=k}^{n-k}\binom{n-1}ta^{t(n-t)}$$
$$\leq(k-1)\binom{n}{k-1}a^{n-1}+\sum\_{t=k}^{n-k}\binom{n-1}ta^{t(n-t)}$$
$$\leq(k-1)\binom{n}{k-1}a^{n-1}+a^{k(n-k)}\sum\_{t=k}^{n-k}\binom{n-1}t$$
$$\leq(k-1)\binom{n}{k-1}a^{n-1}+a^{k(n-k)}2^{n-1}.$$
It remains to observe that (i) $(k-1)\binom{n}{k-1}$ is a polynomial in $n$ while $a^{n-1}$ is a decaying exponentially, so 
$$(k-1)\binom{n}{k-1}a^{n-1}\rightarrow 0 \qquad \text{as $n\rightarrow\infty$}.$$
Also that (ii) since $a^k2<1$, we can say
$$a^{k(n-k)}2^{n-1}=a^{k(1-k)}\cdot(2a^k)^{n-1}\rightarrow0 \qquad \text{as $n\rightarrow\infty$}.$$
We've just shown that $\sum\_{t=1}^{n-1}\rightarrow0$ for each $0<a<\sqrt[k]{\frac12}$. To complete the proof, increase $k\rightarrow\infty$ freely so that the interval $0<a<1$ is exhausted. 


